Brian Bi
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Section 7.6. Normalizers

Exercise 7.6.1 They are conjugate by the unit antidiagonal matrix $$J_n$$, for example, for $$n = 3$$, $J_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$ Verification is left as an exercise.

Exercise 7.6.2 Fix $$n$$; we will assume all matrices are $$n \times n$$ in what follows. We will use the fact the group $$U$$ of upper triangular matrices with all diagonal entries equal to one is generated by the elementary matrices of the first type (1.2.3) that are upper triangular. Let $$E(i, j, c)$$ be the elementary matrix of the first kind where $$E_{ij} = c$$ and $$j > i$$. Let $$A$$ be some other matrix. Then $$E(i, j, c)A$$ is $$A$$ with $$c$$ times the $$j$$th row added to the $$i$$th row, while $$AE(i, j, c)$$ is $$A$$ with $$c$$ times the $$i$$th column added to the $$j$$th column. Therefore, in general, left-multiplication by an element of $$U$$ is equivalent to some sequence of elementary row operations of the first kind that propagate upward and vice versa while right-multiplication by an element of $$U$$ is equivalent to some sequence of elementary column operations of the first kind that propagate rightward and vice versa. This gives us the intuition necessary to solve the problem.

Suppose $$A$$ is invertible but not upper triangular. Let $$r$$ be the bottommost row of $$A$$ that contains a nonzero entry to the left of the diagonal, call it $$A_{rc}$$ (where $$r > c$$). Then $$A E(c, r, 1)$$ differs from $$A$$ at the diagonal entry $$(r, r)$$. There is no sequence of upward elementary row operations of the first kind that can reproduce $$A E(c, r, 1)$$ starting from $$A$$, since by the assumption of the maximality of $$r$$, all entries below $$A_{rr}$$ vanish (and thus, such operations can't modify $$A_{rr}$$). Therefore the equation $$A E(c, r, 1) = MA$$ has no solution with $$M \in U$$. This establishes $$A E(c, r, 1) A^{-1} \notin U$$. Therefore $$A \notin N(U)$$. Since this was true for all $$A \in GL_n(\mathbb{C}) \setminus B$$, we conclude $$N(U) \subseteq B$$.

To see that $$N(B) \subseteq B$$ the idea is very similar—we just need to add the fact that $$B$$ is generated by the upper triangular elementary matrices of the first type together with the elementary matrices of the third type. Given $$A \in GL_n(\mathbb{C}) \setminus B$$, choose $$r, c$$ as above and note that $$A E(c, r, 1) = MA$$ has no solution with $$M \in B$$ either since the only allowed type of row operation that can modify either $$A_{rc}$$ or $$A_{rr}$$ is scaling of row $$r$$, but this will never have the same effect as adding $$A_{rc}$$ to $$A_{rr}$$ given that the former is nonzero.

It is evident that $$B \subseteq N(B)$$, therefore $$N(B) = B$$. Also, the product of two upper triangular matrices has diagonal entries given by the products of corresponding diagonal entries in the two factors, so the conjugation of a matrix in $$U$$ by one in $$B$$ not only lies in $$B$$ but also lies in $$U$$ as well. Therefore $$B \subseteq N(U)$$ as well, which establishes $$N(U) = B$$.

Exercise 7.6.3 As in Exercise 7.6.2, we work with fixed $$n$$. Suppose $$A \in GL_n(\mathbb{R})$$. If $$\sigma \in P$$, then $$\sigma A$$ is $$A$$ with its row set permuted, and $$A \sigma$$ is $$A$$ with its column set permuted. If $$A \in N(P)$$, then this is equivalent to saying that for all permutations $$\sigma \in P$$, we can find $$\tau \in P$$ with $$A\sigma = \tau A$$. That is, every matrix obtained by permuting $$A$$'s column set is also obtainable by permuting $$A$$'s row set. Our task is simply to find all invertible matrices with this property.

First, note that if $$A$$ has this property then each of its rows must itself have no more than $$n$$ distinct permutations of its entries. If some row's entries can be permuted in more than $$n$$ ways, at least one of those must be unequal to an existing row of $$A$$. But that permutation of the row's entries can be formed by permuting the columns of $$A$$ and, by assumption, not by permuting the rows of $$A$$. Such $$A$$ therefore lie outside $$N(P)$$.

The only way that a row can have $$\le n$$ distinct permutations of its own elements is if it contains at most two distinct values among its elements, and if it does contain two distinct values, one of those appears exactly once. (For example, there are more than four permutations of $$[1, 2, 3, 3]$$ and more than four permutations of $$[1, 1, 2, 2]$$, but only four permutations of $$[1, 2, 2, 2]$$.)

If $$A \in N(P)$$ does have some row that contains two distinct values, then there are $$n$$ permutations of this row's elements, and each of them can be accomplished by permuting $$A$$'s set of columns, so each must also already occur among the set of $$A$$'s rows. So either all rows of $$A$$ consist of the same two values, the same one of which appears exactly once in each row, or all rows of $$A$$ contain only one distinct value. The latter would imply that all columns of $$A$$ are identical (so $$A$$ is singular) so we do not consider it further.

At this point we know that every $$A \in N(P)$$ contains the same value $$a \in \mathbb{R}$$ once and the same value $$b \in \mathbb{R}$$ $$n-1$$ times. It is also clear that the entry $$a$$ must occur in a different position in each row, so that all $$n$$ possible permutations of each row can occur among $$A$$'s column set. An example is: $A = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 1 \end{pmatrix}$ Such matrices do have the property that every matrix obtained by permuting their column set is also obtainable by permuting their row set, and are invertible provided $$a \neq b$$. Another way of writing this is $N(P) = \{\alpha J + \beta \sigma \mid \sigma \in P, \alpha \in \mathbb{R}, \beta \in \mathbb{R} \setminus \{0\}\}$ where $$J$$ denotes the all-ones matrix.

Exercise 7.6.4 The subgroup $$H$$ must be cyclic since it is of prime order: $$H = \{1, x, x^2, \ldots, x^{p-1}\}$$. Thus, for every $$g \in G$$, either $$gxg^{-1} = x$$ or $$gxg^{-1} = x^k$$ where $$k \in \{2, 3, \ldots, p-1\}$$. Suppose the latter is true. Then $$k$$ generates a subgroup of $$(\mathbb{Z}/p\mathbb{Z})^\times$$, of order greater than 1 but less than $$p$$. $$k$$ therefore has some prime divisor $$q < p$$. Let $$o$$ be the order of $$g$$. Then $$x = g^o x g^{-o} = x^{k^o}$$, where $$k^o$$ is taken in $$(\mathbb{Z}/p\mathbb{Z})^\times$$. But $$o$$ is known to be a product of primes greater than or equal to $$p$$, so it cannot be divisible by the order of $$k$$, which has a prime divisor $$q$$ less than $$p$$. So $$k^o \neq 1$$, that is, $$g^o x g^{-o} \neq x$$. A contradiction has been reached. So it must in fact be the case that $$gxg^{-1} = x$$ for all $$x$$. If all of $$G$$ commutes with $$x$$, then all of $$G$$ commutes with $$H = \langle x \rangle$$ as well. That is, $$H \in Z(G)$$.

Exercise 7.6.5 (This solution is due to Andrej Vuković.) The proof of the first part is by induction. For the base case, let $$|G| = p^2$$. Then $$|H| = p$$. By Proposition 7.3.3, $$G$$ is abelian, therefore $$N(H) = G$$. For the inductive case, suppose the claim holds for $$|G| = p^2, p^3, \ldots, p^{n-1}$$ and suppose $$|G| = p^n$$, and suppose $$N(H) = H$$. Now $$Z(G) \subseteq N(H)$$, so $$Z(G) \subseteq H$$. By Proposition 7.3.1, $$Z(G)$$ is not trivial. Also, $$Z(G)$$ can't be all of $$H$$ because in that case $$N(H)$$ would be all of $$G$$. Let $$G' = G/Z(G)$$ and $$H' = H/Z(G)$$. Then $$G'$$ is a $$p$$-group of order less than $$p^n$$, $$H'$$ is a nontrivial $$p$$-group, and $$H'$$ is smaller than $$G'$$. Every coset of $$Z(G)$$ in $$H$$ is also a coset of $$Z(G)$$ in $$G$$, so $$H'$$ is a proper subgroup of $$G'$$.

Suppose $$g' \in G' \setminus H'$$. Then $$g'$$ is the coset $$[gZ(G)]$$ where $$g \in G \setminus H$$. By assumption that $$N(H) = H$$, there exists some $$h \in H$$ such that $$ghg^{-1} \notin H$$. Let $$h' = [hZ(G)]$$. Then $$ghg^{-1}$$ is a member of the coset $$[gZ(G)][hZ(G)][gZ(G)]^{-1} = g'h'g'^{-1}$$. This implies that this coset is not any of the cosets of $$Z(G)$$ in $$H$$. Since this construction works for all $$g' \in G' \setminus H'$$, it establishes that the normalizer of $$H'$$ in $$G'$$ is $$H'$$. This contradicts the induction hypothesis, so our assumption that $$N(H) = H$$ must have been faulty.

For the second part, suppose $$H$$ is a proper subgroup of $$G$$ with index greater than $$p$$. Then there are two operations we can do in order to extend $$H$$ to a larger proper subgroup of $$G$$:

• If $$H$$ isn't normal, then $$N(H)$$ is smaller than $$G$$, and by the first part, $$N(H)$$ is also larger than $$H$$. So $$N(H)$$ is a larger proper subgroup of $$G$$.
• If $$H$$ is normal, let $$g \in G \setminus H$$ and let $$k$$ be the smallest positive integer such that $$g^{p^k} \in H$$. Let $$g' = g^{p^{k-1}}$$, so that $$g' \notin H$$ but $$g'^p \in H$$. Let $$K = \langle g' \rangle$$ and $$K' = \{1, g', \ldots, g'^{p-1}\}$$. (Note that $$K'$$ is not necessarily a subgroup of $$G$$.) We claim that $$HK = HK'$$. (To see this, let $$h \in H, k = g'^e$$; then use the division algorithm to write $$e = qp + r$$ where $$0 \le r \le p - 1$$, so $$hk = (hg'^{qp}) g'^r$$ where $$hg'^{qp} \in H, g'^r \in K'$$.) Now Proposition 2.11.4(c) tells us that $$HK$$ is a subgroup of $$G$$. Clearly this subgroup is strictly larger than $$H$$. But $$HK = HK'$$ and $$|HK'| \le |H||K'| = p|H| < |G|$$. So in fact this subgroup $$HK$$ is a larger proper subgroup of $$G$$.

We can always keep extending $$H$$ using one of these two operations until we obtain a subgroup $$H' \supseteq H$$ with index $$p$$ in $$G$$. By the first part, $$N(H')$$ is strictly larger than $$H'$$, so it can only be $$G$$ itself. That is, $$H'$$ is normal.

Exercise 7.6.6 (This solution is due to Tom Guo. Yes, I should have been able to come up with this myself, but I got stuck somehow ><) The number of conjugate subgroups of $$H$$ in $$G$$ is $$[G : N(H)] = |G|/|N(H)|$$. But $$|N(H)| \geq |H|$$, so the number of conjugate subgroups is at most $$|G|/|H|$$. However, all conjugate subgroups share the identity element, so the size of the union of these subgroups is at most $$1 + (|H| - 1)|G|/|H| = |G| + 1 - |G|/|H|$$. Since $$|H|$$ is a proper subgroup, this value is less than $$|G|$$, so this proves part (a). Part (b) is obviously equivalent to part (a).