Brian Bi

## Section 7.3. $$p$$-Groups

Exercise 7.3.1 The action of $$G$$ partitions $$S$$ into orbits. Since $$|S|$$ is not divisible by $$p$$, there must be some orbit $$O$$ whose size is not divisible by $$p$$. But by the orbit-stabilizer theorem, $$|O| \divides |G|$$. Therefore $$|O|$$ can only be 1, and the sole element of $$O$$ is therefore fixed by $$G$$.

Exercise 7.3.2 We are given that $$G/Z$$ is cyclic, so there is some coset $$xZ$$ that generates $$G/Z$$. That is, for every $$g \in G$$, there is some $$n$$ such that $$g \in (xZ)^n$$. Now $$(xZ)^n$$ is just the coset $$x^n Z$$, so $$g = x^n z$$ for some $$z \in Z$$. For every $$g' \in G$$, we can likewise write $$g' = x^{n'} z'$$ for some $$n'$$ and some $$z' \in Z$$. By inspection, $$g$$ and $$g'$$ commute. We conclude that $$G$$ is abelian.

Exercise 7.3.3

1. $$|Z|$$ must divide $$p^3$$. It cannot be 1 according to Proposition 7.3.1. It also cannot be $$p^3$$ because we are given that $$G$$ is nonabelian. The case $$|Z| = p^2$$ is ruled out by the Lemma from Section 7.2 exercises since all remaining conjugacy classes (besides those of the central elements) would have to have size $$p$$. Therefore the only possible order of the centre is $$p$$.
2. Since the order of the centre is $$p$$, the Lemma rules out the existence of any conjugacy classes of size $$p^2$$. So all non-central elements belong to a conjugacy class of size $$p$$, for which the centralizer has order $$p^2$$.
3. The class equation can only be $$p \times 1 + (p^2 - 1) \times p$$ by the above reasoning, where $$m \times n$$ denotes $$m$$ conjugacy classes of size $$n$$.

Exercise 7.3.4 All elements of $$G$$ other than the identity must have order 2, 4, or 8. Choose some $$x \in G$$ of maximal order and let $$H = \langle x \rangle$$. The classification is by cases:

1. $$x$$ has order 8. Thus, $$G \cong C_8$$.
2. $$x$$ has order 4. By Exercise 2.8.10, $$H$$ is normal in $$G$$. Then,
1. Suppose there exists $$y \in G \setminus H$$ such that $$[x, y] = 1$$. If $$y^2 = x^2$$, then set $$y' = yx^{-1}$$ so that $$y'^2 = y^2 x^{-2} = x^2 x^{-2} = 1$$ and use $$y'$$ in place of $$y$$. Let $$K = \langle y \rangle$$; now we know that $$H \cap K = \{1\}$$. By Proposition 2.11.4(a), $$HK$$ consists of $$4|K|$$ distinct elements. Therefore $$|K| = 2, HK = G$$, and the elements $$x$$ and $$y$$ generate $$G$$. By Proposition 2.11.4(b), $$G \cong H \times K \cong C_4 \times C_2$$.
2. Otherwise, suppose there exists $$y \in G \setminus H$$ with order 4. Then $$y^3 = y^{-1}$$ also isn't in $$H$$. The element $$xy$$ can't be in $$H$$ or $$\langle y \rangle$$, and neither can $$yx$$, which is also distinct from $$xy$$. So $$G = \{1, x, x^2, x^3, y, y^3, xy, yx\}$$. Now $$yxy^{-1}$$ lies in $$H$$ since $$H$$ is normal, and it must have the same order as $$x$$, so it can only be $$x^3$$. This completely determines the multiplication table of $$G$$. Now the quaternion group $$Q_8$$ has the aforementioned properties with $$x = i, y = j$$. Therefore $$G \cong Q_8$$.
3. Otherwise, all remaining elements of $$G \setminus H$$ have order 2. Let $$y$$ be such an element and let $$K = \{1, y\}$$. By Proposition 2.11.4(a), the set $$HK$$ consists of 8 distinct elements, so $$HK = G$$ as sets. That is, $$G = \{1, x, x^2, x^3, y, xy, x^2 y, x^3 y\}$$. Since $$H$$ is normal and since $$yxy^{-1}$$ has the same order as $$x$$, we must have $$yx = xy^3$$. This completely determines the multiplication table of $$G$$. Now the dihedral group $$D_4$$ has the aforementioned properties with $$x = \rho, y = r$$. Therefore $$G \cong D_4$$.
3. $$x$$ has order 2. Since all non-identity elements have order 2, $$G$$ is abelian. Pick $$y \in G \setminus H$$. Then we have a subgroup $$I = \{1, x, y, xy\}$$ and it is easy to see that $$I \cong C_2 \times C_2$$. Pick $$z \in G \setminus I$$. Then, by Proposition 2.11.4(c), $$I\langle z\rangle$$ is a subgroup of $$G$$ which is larger than $$I$$, therefore it is equal to $$G$$. By Proposition 2.11.4(d), $$G \cong I \times \langle z\rangle \cong C_2 \times C_2 \times C_2$$.

There are thus a total of five isomorphism classes for groups of order 8, namely $$C_8, C_2 \times C_4, C_2 \times C_2 \times C_2, D_4, Q_8$$. (The division into cases makes it clear that these five groups are pairwise nonisomorphic.)