Brian Bi

## Section 7.7. The Sylow Theorems

Exercise 7.7.1 The number of subsets is \begin{align*} \binom{mp^e}{p^e} &= \frac{mp^e (mp^e - 1) \ldots (mp^e - p^e + 2)(mp^e - p^e + 1)}{p^e(p^e - 1)\ldots \cdot 2 \cdot 1} \\ &= \frac{mp^e}{p^e} \frac{mp^e - 1}{p^e - 1} \ldots \frac{mp^e - p^e + 2}{2} \frac{mp^e - p^e + 1}{1} \end{align*} Some of the terms in this product have both numerator and denominator divisible by $$p$$, while the others have neither numerator nor denominator divisible by $$p$$. The terms of the former type can be written in the form $\frac{mp^e - np^f}{p^e - np^f}$ where $$p$$ does not divide $$n$$, and $$f < e$$. Such a fraction can be reduced to the form $\frac{mp^{e-f} - n}{p^{e-f} - n}$ Once this reduction has been performed, all terms in the expansion of $$\binom{mp^e}{p^e}$$ have a numerator and denominator that are not divisible by $$p$$ and have the same remainder upon division by $$p$$. This implies $N = \binom{mp^e}{p^e} = \frac{a}{b}$ where $$a \equiv b \not\equiv 0 \pmod{p}$$. Then, $$a \equiv bN \pmod{p}$$, or $$N \equiv b^{-1}a \equiv 1 \pmod{p}$$.

Exercise 7.7.2 $$H_1$$ is a $$p$$-group, therefore $$H_1$$ is a $$p$$-subgroup of $$G_2$$ and is contained in a Sylow $$p$$-subgroup of $$G_2$$, call it $$H_2$$. Then $$H_2 \cap G_1$$ is a subgroup of both $$H_2$$ and $$G_1$$ that contains $$H_1$$. Because $$H_2$$ is a $$p$$-group, $$H_2 \cap G_1$$ is also a $$p$$-group. But because $$H_1$$ is a Sylow $$p$$-subgroup of $$G_1$$, no $$p$$-group larger than $$H_1$$ can be a subgroup of $$G_1$$. Therefore $$H_1 = H_2 \cap G_1$$.

Exercise 7.7.3 Suppose $$|G| = 20$$. An element of order 5 generates a subgroup of $$G$$ of order 5, which is therefore a Sylow 5-subgroup of $$G$$; conversely, every Sylow 5-subgroup of $$G$$ is a group of order 5 and is therefore generated by its four non-identity elements. By Sylow's third theorem, the number of Sylow 5-subgroups of $$G$$ divides 4 and is congruent to 1 modulo 5; therefore there is exactly one Sylow 5-subgroup. The four non-identity elements of this Sylow 5-subgroup are therefore the only four elements of $$G$$ that have order 5.

Exercise 7.7.4 We note here that a group of order $$p^e$$, where $$e \ge 2$$, is not normal. To see this, find an element of order $$p$$ and use Exercise 7.6.5 to conclude that the subgroup it generates is contained in a normal subgroup of index $$p$$. We therefore assume that $$p \neq q$$ in (a) and (b).

1. Without loss of generality, suppose $$p > q$$. Then the number of Sylow $$p$$-subgroups divides $$q$$ and is congruent to 1 modulo $$p$$; this number can only be 1. Since there is only one Sylow $$p$$-subgroup, it is normal. Therefore the group of order $$pq$$ is not simple.
2. Let $$G$$ have order $$p^2 q$$. The number of Sylow $$q$$-subgroups of $$G$$ is either 1, $$p$$, or $$p^2$$.
• Case 1: There is one Sylow $$q$$-subgroup of $$G$$. Then this is a normal subgroup of $$G$$, and $$G$$ is not simple.
• Case 2: There are $$p$$ Sylow $$q$$-subgroups. Then $$p \equiv 1 \pmod{q}$$, therefore $$p > q$$. This implies that $$q \not\equiv 1 \pmod{p}$$, so the number of Sylow $$p$$-subgroups of $$G$$ is 1. The unique Sylow $$p$$-subgroup is therefore normal, and $$G$$ is not simple.
• Case 3: There are $$p^2$$ Sylow $$q$$-subgroups. Every pair of such subgroups has trivial intersection, so the total number of elements of order $$q$$ is $$p^2(q-1)$$. This only leaves $$p^2$$ elements of order not equal to $$q$$. This implies that there is only one Sylow $$p$$-subgroup of $$G$$, which consists of all $$p^2$$ of those elements. This unique Sylow $$p$$-subgroup is then normal, and $$G$$ is not simple.

Exercise 7.7.5

1. The group $$D_{10}$$ has order 20, so we seek a subgroup of order 4. The subgroup $$\{1, \rho^5, r, \rho^5 r\}$$ is such a subgroup.
2. The group $$T$$ has order 12, so we seek a subgroup of order 4. The subgroup consisting of the identity and the three 180-degree rotations is such a subgroup. (This corresponds to the normal Klein four-subgroup of the alternating group $$A_4$$.)
3. The group $$O$$ has order 24, so we seek a subgroup of order 8. The stabilizer of some set of size 3 on which this group acts transitively would be such a subgroup. For example, the stablizer of a pair of opposite faces of a cube, $$f, f'$$. This subgroup consists of the identity, the three rotations around the axis joining the centres of the two faces, the two 180-degree face rotations that exchange the two faces in question (associated with the two pairs of opposite faces other than $$f, f'$$), and the two 180-degree edge rotations that exchange the two faces in question (associated with the two pairs of opposite edges that belong to neither of $$f, f'$$). We have already classified all groups of order 8 in Exercise 7.3.4. This particular group has 5 elements of order 2 and 2 elements of order 4, so it must be isomorphic to $$D_4$$.
4. The group $$I$$ has order 60, so we seek a subgroup of order 4. We found such a subgroup in Exercise 7.4.3.

Exercise 7.7.6 According to Proposition 7.7.7(c), such a group has an element $$x$$ of order 7 and an element $$y$$ of order 3, and these elements satisfy $$yxy^{-1} = x^2$$. Now the only elements of $$S_7$$ that have order 7 are the 7-cycles, so, without loss of generality, let $$x = (1\ 2\ 3\ 4\ 5\ 6\ 7)$$, so that $$x^2 = (1\ 3\ 5\ 7\ 2\ 4\ 6)$$. We must find $$y$$ such that $$y$$ relabels $$x$$ to $$x^2$$ and has order 3. The obvious relabelling from the way $$x$$ and $$x^2$$ are written, above, is $$y = (2\ 3\ 5)(4\ 7\ 6)$$, which indeed has order 3. Since we have the rule $$yx = x^2 y$$, the elements $$x^i y^j$$ with $$i$$ from 0 to 6 and $$j$$ from 0 to 2 form a subgroup of $$S_7$$ of order at most 21. This subgroup contains both $$\langle x \rangle$$ and $$\langle y \rangle$$ as subgroups, so its order is divisible by 21. Therefore its order is exactly 21.

Exercise 7.7.7 The group $$H$$ has order $$p$$, so $$S$$ is decomposed into orbits of size 1 or $$p$$. Since $$|S| \cong 1 \pmod{p}$$, there is at least one orbit of size 1. Obviously $$H$$ itself (as an element of $$S)$$ is a fixed point. As in the proof of Sylow's third theorem given in the text, we can show that $$H$$ is the only element of $$S$$ that is fixed by conjugation by $$H$$. Namely, suppose $$H'$$ is fixed by $$H$$. Then $$H'$$ and $$H$$ are both Sylow $$p$$-subgroups of $$N(H')$$. But $$H'$$ is a normal subgroup of $$N(H')$$, so it is the only Sylow $$p$$-subgroup of $$N(H')$$. Therefore $$H = H'$$. This shows that the action of $$H$$ on $$S$$ decomposes $$S$$ into a single fixed point and some number of orbits of order $$p$$.

Exercise 7.7.8 The order of $$GL_n(\mathbb{F}_p)$$ is $$\prod_{i=0}^p (p^n - p^i)$$. To see this, observe that it's sufficient and necessary for the $$n$$ columns to be linearly independent. For the first column we can choose any of $$p^n - 1$$ nonzero vectors. After we have chosen the first $$i$$ columns to be linearly independent, there are exactly $$p^i$$ distinct linear combinations of them, so we have $$p^n - p^i$$ choices for the next column. Note that $$p$$ divides this product $$1 + 2 + \ldots + n-1 = n(n-1)/2$$ times, so a Sylow $$p$$-subgroup has order $$p^{n(n-1)/2}$$. The upper unitriangular group $$U$$ is such a subgroup: an upper triangular matrix with each diagonal element equal to 1 will be invertible regardless of the values of the entries above the diagonal, so there are $$n(n-1)/2$$ entries that can be chosen freely. We argued in Exercise 7.6.2 that the normalizer of $$U$$ is the subgroup $$B$$ of all invertible upper triangular matrices; nothing in that proof depended on the choice of the field. The order of $$B$$ is $$(p-1)^n p^{n(n-1)/2}$$ since we can freely choose all entries above the diagonal whereas diagonal entries may be freely chosen as long as they are all nonzero. The number of Sylow $$p$$-subgroups of $$GL_n(\mathbb{F}_p)$$ is then $$|GL_n(\mathbb{F}_p)|/|B|$$, which is $s = \prod_{i=1}^n \frac{p^i - 1}{p-1}$

Exercise 7.7.9

1. A group of order 33 has a Sylow 11-subgroup $$H$$ of order 11, which is generated by some $$x \in G$$, and a Sylow 3-subgroup $$K$$, which is generated by some $$y \in G$$. By Sylow's third theorem, the number of Sylow 11-subgroups must be 1, and the number of Sylow 3-subgroups must also be 1. So both $$H$$ and $$K$$ are normal. As in Proposition 7.7.7(a), it is then clear that $$G \cong H \times K$$. So all groups of order 33 are isomorphic to $$C_{11} \times C_3$$, which is a cyclic group. In general, if $$p$$ and $$q$$ are primes, $$q > p$$, and $$p$$ does not divide $$q - 1$$, then there is only one isomorphism class of groups of order $$pq$$.
2. There is a Sylow 3-subgroup $$H$$, which has order 9. Since this subgroup has index 2, it is normal; there are no other Sylow 3-subgroups. Let $$K$$ be a Sylow 2-subgroup; it is clear that $$G = HK$$. According to Corollary 7.3.4, $$H$$ is either cyclic or the product of two cyclic subgroups of order 3.
• Case 1: $$H$$ is cyclic. Let $$x$$ be a generator of $$H$$ and let $$y$$ be a generator of $$K$$. By normality of $$H$$, $$yxy^{-1} = x^i$$ for some $$i$$. Since $$y^2 = 1$$, it follows that iterating conjugation by $$y$$ a second time yields back $$x$$. So $$i^2 \equiv 1 \pmod{9}$$. Therefore $$i = 1$$ or $$i = 8$$.
• Case 1a: $$i = 1$$. By Theorem 2.11.4, $$G \cong C_2 \times C_9$$.
• Case 1b: $$i = 8$$. The relations $$x^9 = 1, y^2 = 1, yx = x^{-1}y$$ are the defining relations of the dihedral group $$D_9$$. So $$G \cong D_9$$. Note that $$D_9$$ is nonabelian and therefore clearly not isomorphic to $$C_2 \times C_9$$.
• Case 2: $$H$$ is not cyclic. By Corollary 7.3.4, $$H \cong C_3 \times C_3$$. Let $$x, y$$ generate $$H$$, so that $$H = \{x^i y^j \mid i, j \in \{0, 1, 2\}\}$$. Let $$z$$ be a generator of $$K$$. Write $$x' = zxz^{-1}$$ and $$y' = zyz^{-1}$$. Then $$x' = x^a y^c, y' = x^b y^d$$ for some unique $$a, b, c, d \in \mathbb{F}_3^2$$. We can write this as follows: $z \begin{pmatrix} x \\ y \end{pmatrix} z^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$ Because $$x$$ and $$y$$ commute, a second conjugation is equivalent to multiplying by this matrix a second time. But this must restore $$x'$$ to $$x$$ and $$y'$$ to $$y$$. Therefore $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ Call the matrix on the left-hand side $$M$$. If we consider the ground field to be $$\overline{\mathbb{F}_3}$$ (the algebraic closure of $$\mathbb{F}_3$$) then $$M$$ has a Jordan normal form whose square must be an identity matrix. Therefore the eigenvalues must be $$\pm 1$$. The nondiagonalizable cases can be ruled out, as the squares of $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$ and $$\begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}$$ are not the identity matrix. So one of the following three cases occurs:
• Case 2a: $$M$$ has the eigenvalue 1 with a geometric multiplicity of 2. That is, $$M$$ is the identity matrix. So $$z$$ commutes with $$H$$. By Proposition 2.11.4, $$G \cong C_2 \times C_3 \times C_3$$.
• Case 2b: $$M$$ has the eigenvalue 2 with a geometric multiplicity of 2. That is, $$M = 2I$$. This completely determines the multiplication table of $$G$$. This group is the generalized dihedral group corresponding to $$C_3 \times C_3$$ and exists by construction.
• Case 2c: $$M$$ has both 1 and 2 as eigenvalues. Then $$M$$ has two eigenvectors in $$\mathbb{F}_3^2$$, call them $$(a', c')$$ with eigenvalue 1 and $$(b', d')$$ with eigenvalue 2. Replace $$x$$ and $$y$$ with $$x^{a'} y^{c'}$$ and $$x^{b'} y^{d'}$$, respectively, so that $$zxz^{-1} = x$$ and $$zyz^{-1} = y^2$$. Then $$y, z$$ generate a subgroup of $$G$$ that is isomorphic to $$S_3$$ and which commutes with $$\langle x \rangle$$, so $$G \cong S_3 \times C_3$$.
The groups in cases 2b and 2c are nonabelian, so they are definitely not isomorphic to the group $$C_2 \times C_3 \times C_3$$. These two groups are also not isomorphic to each other, as the generalized dihedral group $$\operatorname{Dih}(C_3 \times C_3)$$ contains 9 elements of order 2, while $$S_3 \times C_3$$ only contains 3 elements of order 2.
Since there is only one Sylow 3-subgroup, the groups described in Case 1 are not isomorphic to the groups in Case 2. There are therefore exactly five isomorphism classes for groups of order 18, namely $$C_2 \times C_9$$ (which is also $$C_{18}$$), $$D_9$$, $$C_2 \times C_3 \times C_3$$, $$\operatorname{Dih}(C_3 \times C_3)$$, and $$S_3 \times C_3$$.
3. The number of Sylow 5-subgroups divides 4 and is congruent to 1 modulo 5, so there is exactly one Sylow 5-subgroup, call it $$H$$, with generator $$x$$, and it is normal. Let $$K$$ denote a Sylow 2-subgroup of $$G$$, with order 4. It is clear that $$G = HK$$. The group $$K$$ is isomorphic to either $$C_4$$ or $$V_4$$.

• Case 1: $$K \cong C_4$$. Let $$y$$ be a generator of $$K$$. Since $$H$$ is normal, $$yxy^{-1} = x^i$$ for some $$i$$.
• Case 1a: $$i = 1$$. By Proposition 2.11.4, $$G$$ is simply the group $$H \times K \cong C_5 \times C_4$$, which is also the cyclic group $$C_{20}$$.
• Case 1b: $$i = 2$$ or $$i = 3$$. Note that if $$i = 3$$ then $$y^3 x y^{-3} = x^2$$, so we can simply replace $$y$$ with $$y^2$$. We therefore assume without loss of generality that $$i = 2$$. The multiplication table of $$G$$ is now completely determined. The isomorphism class of $$G$$ is called the general affine group $$GA(1, 5)$$. It can be represented as a subgroup of $$S_5$$ with $$x = (1\ 2\ 3\ 4\ 5), y = (2\ 3\ 5\ 4)$$, which satisfy the defining relations and therefore generate a group of order at most 20, which is divisible by both 4 and 5 (and is therefore exactly 20).
• Case 1c: $$i = 4$$. Then $$y^2$$ commutes with $$x$$. Therefore $$x$$ and $$y^2$$ generate a subgroup of $$G$$ of order 10 which is isomorphic to $$C_2 \times C_5$$ or alternatively the cyclic group of order 10. Let $$x'$$ be a generator of this subgroup, so that $$y^2 = x'^5$$. Call this subgroup $$I$$. It is clear from Proposition 2.11.4 that $$\langle y\rangle I = G$$. Since $$I$$ has index 2, it is normal in $$G$$. Now $$y$$ is known not to commute with $$x'$$ (if it did, it would also commute with $$x$$, which is a power of $$x'$$ by assumption). But conjugation by $$y$$ twice is conjugation by $$y^2 = x'^5$$, so if $$yx'y^{-1} = x'^i$$, then $$i^2 \equiv 1 \pmod{10}$$. The only possibility is that $$yx'y^{-1} = x^{-1}$$. This completely determines the multiplication table for the group. The isomorphism class of $$G$$ is called the dicyclic group $$\operatorname{Dic}_5$$. By trial and error, we can represent this group as a subgroup of $$S_9$$ with $$x' = (1\ 2\ 3\ 4\ 5)(6\ 8)(7\ 9), y = (2\ 5)(3\ 4)(6\ 7\ 8\ 9)$$, where $$x'^5 = y^2, x'^{10} = y^4 = 1, yx'y^{-1} = x'^{-1}$$, so the defining relations are satisfied and $$x'$$ and $$y$$ generate a group of order at most 20, which is divisible by 4 and 5, and is therefore exactly 20.
• Case 2: $$K \cong V_4$$. Let $$y, z$$ be two non-identity elements of $$K$$. Each of these must act by conjugation on $$x$$ by either sending it to itself or to $$x^4$$.
• Case 2a: $$y, z$$ both commute with $$x$$. Then by Proposition 2.11.4, $$G \cong H \times K \cong C_2 \times C_2 \times C_5 \cong C_2 \times C_{10}$$.
• Case 2b: Either $$y$$ or $$z$$ commutes with $$x$$, but the other doesn't. Without loss of generality, suppose $$y$$ commutes with $$x$$. Then $$y$$ and $$x$$ generate a subgroup $$I$$ of order 10 which is isomorphic to $$C_{10}$$ and which is known to be normal since it has index 2 in $$G$$. Let $$x'$$ be a generator of $$I$$, for example, $$x' = xy$$. Since $$z$$ doesn't commute with $$x'$$, the only possibility is that $$zx'z^{-1} = x'^{-1}$$ (as in Case 1c). Consequently $$G \cong D_{10}$$.
• Case 2c: Neither of $$y, z$$ commute with $$x$$, that is, $$yxy^{-1} = zxz^{-1} = x^4$$. Then $$yz$$ is an element of order 2 that does commute with $$x$$. Replace $$y$$ with $$yz$$ and apply Case 2b.

There are therefore five isomorphism classes of groups of order 20, namely $$C_{20}, GA(1, 5), \operatorname{Dic}_5, C_2 \times C_{10}, D_{10}$$. To see that these are really pairwise nonisomorphic, first observe that Case 1 and Case 2 are disjoint since all Sylow 2-subgroups are conjugate to each other (and hence isomorphic as groups); if $$G$$ contains a cyclic subgroup of order 4 then it does not contain a Klein four-subgroup, and vice versa. Within Case 2, the groups $$C_2 \times C_{10}$$ and $$D_{10}$$ are nonisomorphic as the former is abelian and the latter is not. Likewise $$C_{20}$$ is nonisomorphic to the other two groups in Case 1.

To see that $$GA(1, 5)$$ and $$\operatorname{Dic}_5$$ are not isomorphic to each other, we will again use the fact that all Sylow 2-subgroups are conjugate to each other. In $$\operatorname{Dic}_5$$, every element $$y'$$ of order 4 generates a subgroup that is conjugate to $$K$$. Therefore $$y'^2$$ is conjugate to some element of $$K$$ of order 2, which can only be $$y^2$$. But $$y^2 = x'^5$$, and the $$C_{10}$$ subgroup is normal, and the only conjugate of 5 in $$C_{10}$$ is 5 itself, therefore $$y'^2 = x'^5$$ for every element $$y'$$ of order 4 in $$\operatorname{Dic}_5$$. This establishes that every element of order 4 in $$\operatorname{Dic}_5$$ has a square that commutes with the unique Sylow 5-subgroup $$H$$. Meanwhile, in $$GA(1, 5)$$, there is an element of order 4 whose square doesn't commute with $$x$$, but instead $$y^2 x y^{-2} = x^4$$. So these two groups can't be isomorphic.

4. By Sylow's third theorem, the number of Sylow 5-subgroups is either 1 or 6, and the number of Sylow 3-subgroups is either 1 or 10. If there are 6 Sylow 5-subgroups, then they contain among them 24 distinct elements of order 5. If there are 10 Sylow 3-subgroups, then they contain among them 20 distinct elements of order 3. Clearly both of these can't be true in a group of order 30. So at least one of the two groups is normal. Let $$H = \langle x \rangle$$ be a subgroup of order 5, and $$K = \langle y \rangle$$ be a subgroup of order 3.

• Case 1: $$H$$ is normal. Then $$yxy^{-1} = x^i$$ for some $$i \in \{1, 2, 3, 4\}$$. But $$y$$ has order 3, so it must be that $$i^3 \equiv 1 \pmod{5}$$ so that we get $$y^3 x y^{-3} = x$$. The only possibility is then $$i = 1$$. That is, $$H$$ and $$K$$ commute, and by Proposition 2.11.4, $$x$$ and $$y$$ generate a subgroup of $$G$$ isomorphic to $$C_5 \times C_3$$.
• Case 2: $$K$$ is normal. Then $$xyx^{-1} = y^i$$ for $$i = 1$$ or $$i = 2$$. Again, we can rule out the $$i = 2$$ case since $$2^5 \not\equiv 1 \pmod{3}$$. So again, $$H$$ and $$K$$ commute and generate a subgroup of $$G$$ isomorphic to $$C_5 \times C_3$$.

So we know for sure that there is a subgroup $$I \subseteq G$$ that is isomorphic to $$C_5 \times C_3$$, with generators $$x$$ and $$y$$ of orders 5 and 3, respectively. Let $$z$$ have order 2 in $$G$$. It is clear that $$\langle z \rangle I = G$$. The conjugate $$zxz^{-1}$$ can be either $$x$$ or $$x^4$$, while $$zyz^{-1}$$ can be either $$x$$ or $$x^2$$.

• Case 1: $$z$$ commutes with both $$x$$ and $$y$$. Then $$\langle z \rangle$$ commutes with $$I$$. By Proposition 2.11.4, $$G \cong C_2 \times C_3 \times C_5$$.
• Case 2: $$z$$ commutes with $$x$$ but not with $$y$$, so that $$zyz^{-1} = y^2$$. Then $$y, z$$ generate a subgroup of $$G$$ that is isomorphic to $$S_3$$, and elements of this subgroup commute with $$x$$. By Proposition 2.11.4, $$G \cong S_3 \times C_5$$.
• Case 3: $$z$$ commutes with $$y$$ but not with $$x$$, so that $$zxz^{-1} = z^4$$. Then $$x, z$$ generate a subgroup of $$G$$ that is isomorphic to $$D_5$$, and elements of this subgroup commute with $$y$$. By Proposition 2.11.4, $$G \cong D_5 \times C_3$$.
• Case 4: $$z$$ commutes with neither $$x$$ nor $$y$$. Note that $$xy$$ is a generator of $$I$$, and that $$zxyz^{-1} = zxz^{-1}zyz^{-1} = x^4 y^2 = (xy)^{-1}$$, that is, conjugation by $$z$$ maps each of the elements of the cyclic group $$I$$ to its inverse. Therefore $$G \cong D_{15}$$.

So there are four isomorphism classes of groups of order 30, namely $$C_2 \times C_3 \times C_5$$ (which is also the cyclic group of order 30), and three nonabelian groups, namely $$S_3 \times C_5, D_5 \times C_3, D_{15}$$. These groups respectively have 3, 5, and 15 elements of order 2, so they are pairwise nonisomorphic.

Exercise 7.7.10 In Exercise 7.7.4 we ruled out some possible orders of groups from being simple. The remaining composite orders less than 60 are 24, 36, 40, 42, 48, 54, and 56. We tackle these cases separately, but there are some overlaps. We rely on the use of Sylow's third theorem to narrow the possible numbers of Sylow $$p$$-subgroups.

A group of order $$24 = 8 \times 3$$ contains either 1 or 4 Sylow 3-subgroups. If it's the former, then that subgroup is normal, so the group is not simple. If it's the latter, then the group acts by conjugation transitively on the set of four Sylow 3-subgroups, defining a nontrivial homomorphism $$\varphi : G \to S_4$$. If the kernel is trivial, then $$G \cong S_4$$, which is not simple. If the kernel is nontrivial, the kernel is a proper normal subgroup.

A group of order $$36 = 4 \times 3^2$$ contains either 1 or 4 Sylow 3-subgroups. If it's the former, then that subgroup is normal. If it's the latter, then the group acts by conjugation transitively on the set of four Sylow 3-subgroups, defining a nontrivial homomorphism $$\varphi : G \to S_4$$. Since $$G$$ has order greater than 24, this homomorphism has nontrivial kernel, which is then a proper normal subgroup.

A group of order $$40 = 8 \times 5$$ must have exactly one Sylow 5-subgroup, so that subgroup is normal.

A group of order $$42 = 6 \times 7$$ must have exactly one Sylow 7-subgroup, so that subgroup is normal.

A group of order $$48 = 16 \times 3$$ contains 1, 4, or 16 Sylow 3-subgroups. If the number is 1 or 4, then a similar argument to the case of order 36 applies. If it's 16, then the 16 Sylow 3-subgroups contain among them a total of 32 distinct elements of order 3, leaving only 16 elements that could be part of Sylow 2-subgroups. But a Sylow 2-subgroup would have order 16, so there is only one such subgroup, so it is normal.

A group of order $$54 = 2 \times 3^3$$ must have exactly one Sylow 3-subgroup, so that subgroup is normal.

A group of order $$56 = 8 \times 7$$ contains either 1 or 8 Sylow 7-subgroups. If it's the former, then that subgroup is normal. If it's the latter, then the 8 Sylow 7-subgroups contain among them a total of 48 distinct elements of order 7, leaving only 8 elements that could be part of Sylow 2-subgroups. But a Sylow 2-subgroup would have order 8, so there is only one such subgroup, so it is normal.