Exponent of finite groups (2)

Posted: November 9, 2021 in Basic Algebra, Groups
Tags: dihedral group, exponent of a group, finite abelian group, finite commutative ring, fundamental theorem for finite abelian groups, general linear group, Heisenberg group, Sylow subgroup, symmetric group

See the first part of this post here. Now we find the exponent of some finite groups.

Example 1. Let $G$ be a finite group. Show that if $G$ is cyclic or $|G|$ is square-free, then $\exp(G)=|G|.$

Solution. We already showed in Remark v), in the first of this post, that if $G$ is cyclic, then $\exp(G)=|G|.$ Now suppose that $|G|=p_1 \cdots p_n,$ where $p_1, \cdots, p_n$ are distinct primes. By Remarks i), iii), in the first part of this post, $\exp(G) \mid |G|$ and $p_i \mid \exp(G)$ for all $i.$ So $|G|=p_1 \cdots p_n \mid \exp(G)$ and the result follows. $\Box$

Example 2. Let $G$ be a finite abelian group. Find $\exp(G)$ and show that there exists $x \in G$ such that $\exp(G)=o(x).$

Solution. By the fundamental theorem for finite abelian groups, there exists positive integers $n, d_1, \cdots, d_n$ such that $d_i \mid d_{i+1}$ for all $1 \le i \le n-1$ and $G \cong \bigoplus_{i=1}^n\mathbb{Z}_{d_i}.$ Thus by Remarks v), viii), in the first part of this post

$\exp(G)=\text{lcm}(\exp(\mathbb{Z}_{d_1}), \cdots , \exp(\mathbb{Z}_{d_n})=\text{lcm}(d_1, \cdots, d_n)=d_n.$

Note that if $x=(0, \cdots, 0, 1) \in \bigoplus_{i=1}^n\mathbb{Z}_{d_i},$ then $o(x)=d_n=\exp(G). \ \Box$

Example 3. Let $D_{2n}, \ n \ge 1,$ be the dihedral group of order $2n.$ Show that $\exp(D_{2n})=\text{lcm}(2,n).$

Solution. Recall that

$D_{2n}=\langle a,b \ | \ \ a^2=(ab)^2=b^n=1\rangle.$

So an element of $D_{2n}$ is in the form $a^ib^j, \ 0 \le i \le 1, \ 0 \le j \le n-1.$ Now, we have $o(ab^j)=2$ and $\displaystyle o(b^j)=\frac{o(b)}{\gcd(j,n)} \mid o(b)=n,$ for all $0 \le j \le n-1.$ Thus, by Remark ii) in the first part of this post,

$\begin{aligned}\exp(D_{2n})=\text{lcm}(o(x): \ x \in D_{2n})=\text{lcm}(o(a^ib^j): \ 0 \le i \le 1, \ 0 \le j \le n-1\}=\text{lcm}(2,n). \ \Box \end{aligned}$

Example 4. Show that $\exp(S_n)=\text{lcm}(1,2, \cdots,n),$ where, as always, $S_n$ is the group of permutations of a set of $n$ objects.

Solution. Let $m:=\text{lcm}(1,2, \cdots,n)$ and suppose that $\sigma_k \in S_n, \ 1 \le k \le n,$ is a cycle of length $k.$ Then $o(\sigma_k)=k,$ and so $k \mid \exp(S_n)$ for all $1 \le k \le n.$ Thus

$m \mid \exp(S_n). \ \ \ \ \ \ \ \ \ \ \ (1)$

Now, let $\sigma \in S_n,$ and let $\sigma=\sigma_1\sigma_2 \cdots \sigma_k$ be the decomposition of $\sigma$ into disjoint cycles. Clearly $o(\sigma_i) \le n$ and thus $\sigma_i^m=1$ for all $i.$ Hence, since $\sigma_i\sigma_j=\sigma_j\sigma_i$ for all $i,j,$ we have $\sigma^m=\sigma_1^m \sigma_2^m \cdots \sigma_k^m=1$ and so

$\exp(S_n) \mid m. \ \ \ \ \ \ \ \ \ \ \ (2)$

The result now follows from $(1),(2). \ \Box$

Example 5. Let $R$ be a finite commutative ring with $1,$ and let $G:=H(R),$ the Heisenberg group over $R.$ Let $n:=\text{char}(R),$ the characteristic of $R.$ Show that $\exp(G)=\gcd(2,n)n.$

Solution. Let

$\displaystyle x=\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}, \ \ \ \ \ a,b,c \in R,$

be an element of $G.$ A simple induction shows that for all positive integers $m$ we have

$\displaystyle x^m= \begin{pmatrix}1 & ma & mb + \frac{m(m-1)}{2}ac \\ 0 & 1 & mc \\ 0 & 0 & 1 \end{pmatrix}. \ \ \ \ \ \ \ \ \ (1)$

Thus, since $n=\text{char}(R),$ we have $x^{\gcd(2,n)n}=I,$ the identity matrix, and so

$\exp(G) \mid \gcd(2,n)n, \ \ \ \ \ \ \ \ \ \ \ \ (2)$

by Remark i) in the first part of this post. Now, let

$\displaystyle x=\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \in G$

and suppose that $m:=o(x).$ Then $x^m=I$ and $(1)$ give $\displaystyle m1_R=\frac{m(m-1)}{2}1_R=0$ and so there exist positive integers $k,\ell$ such that

$\displaystyle m=kn, \ \ \ \ \frac{m(m-1)}{2}=\ell n,$

which give $\displaystyle \frac{k(kn-1)}{2}=\ell.$ Thus, since $o(x)=kn,$ we just need to find the smallest integer $k \ge 1$ for which $\displaystyle \ell$ is an integer. Well, if $n$ is even, then $\ell$ is an integer if and only if $2 \mid k$ and so $k=2,$ which gives $o(x)=2n.$ If $n$ is odd, then $\ell$ is an integer for $k=1$ and so $o(x)=n.$ Thus $o(x)=\gcd(2,n)n$ and so

$\gcd(2,n)n=o(x) \mid \exp(G). \ \ \ \ \ \ \ \ \ \ (3)$

The result now follows from $(2),(3). \ \Box$

Example 6. Let $G:=\text{GL}(3,\mathbb{Z}_2),$ the group of $3 \times 3$ invertible matrices with entries from $\mathbb{Z}_2.$ Show that $|G|=168$ and $\exp(G)=84.$

Solution. Note that $G$ is the group of units of $M_3(\mathbb{Z}_2)$ and so, by Problem 3 in this post,

$|G|=168=2^3 \times 3 \times 7.$

Let $P_1=H(\mathbb{Z}_2),$ the Heisenberg group over $\mathbb{Z}_2.$ Clearly $|P_1|=8$ and so $P_1$ is a Sylow $2$ -subgroup of $G.$ Let $P_2,P_3$ be, respectively, a Sylow $3$ -subgroup and a Sylow $7$ -subgroup of $G.$ By Example 5, $\exp(P_1)=4$ and by Example 1, $\exp(P_2)=|P_2|=3, \ \exp(P_3)=|P_3|=7.$ Hence, by the Proposition in the first part of this post,