## Tag Archives: order (group element)

### The order of the Frobenius map on a finite field

Let $\varphi$ denote the Frobenius map $x \mapsto x^p$ on $\mathbb{F}_{p^n}$. Prove that $\varphi$ is an automorphism and compute its order in $\mathsf{Aut}(\mathbb{F}_{p^n})$.

Recall that $\varphi$ is a homomorphism. Moreover, if $\alpha \in \mathsf{ker}\ \varphi$, then $\alpha^p = 0$. Since fields contain no nontrivial zero divisors, we have $\alpha = 0$ (using induction if you want). So the kernel of $\varphi$ is trivial, and thus $\varphi$ is injective. Since $\mathbb{F}_{p^n}$ is finite, $\varphi$ is surjective, and so is a field isomorphism.

Next, we claim that $\varphi^t(\alpha) = \alpha^{p^t}$ for all $\alpha$ and all $t \geq 1$, and will show this by induction. The base case certainly holds, and if $\varphi^t(\alpha) = \alpha^{p^t}$, then $\varphi^{t+1}(\alpha) = \varphi(\varphi^t(\alpha)) = \varphi(\alpha^{p^t})$ $= (\alpha^{p^t})^p$ $= \alpha^{p^{t+1}}$ as desired.

Now $\varphi^n(\alpha) = \alpha^{p^n} = \alpha$, since the elements of $\mathbb{F}_{p^n}$ are precisely the roots of $x^{p^n}-x$. So we have $\varphi^n = 1$.

If $\varphi^t = 1$, then we have $\alpha^{p^t} - \alpha = 0$ for all $\alpha$, so that each $\alpha$ is a root of $x^{p^t}-x$. So $x^{p^n-1}-1$ divides $x^{p^t-1}-1$, and so $p^n-1$ divides $p^t-1$ (by this previous exercise) and then $n$ divides $t$ (by this previous exercise). In particular, $n \leq t$.

So $n$ is the order of $\varphi$ in $\mathsf{Aut}(\mathbb{F}_{p^n})$.

### There are no 3×3 matrices over QQ of multiplicative order 8

Show that there do not exist $3 \times 3$ matrices $A$ over $\mathbb{Q}$ such that $A^8 = I$ and $A^4 \neq I$.

If $A$ is such a matrix, then the minimal polynomial of $A$ divides $x^8-1 = (x^4-1)(x^4+1)$ but not $x^4-1$; so the minimal polynomial of $A$ divides $x^4+1$. Since $A$ has dimension 3, the characteristic polynomial of $A$ has degree 3, so the minimal polynomial has degree at most 3. Note that $(x+1)^4+1 = x^4 + 4x^3 + 6x^2 + 4x + 2$ is Eisenstein at 2, and so $x^4+1$ is irreducible over $\mathbb{Q}$. (See Proposition 13 on page 309 in D&F, and Example 3 on page 310.) So we have a contradiction- no divisor of $x^4+1$ over $\mathbb{Q}$ can have degree between 1 and 3. So no such matrix $A$ exists.

### Every nonidentity element in a free group has infinite order

Prove that every nonidentity element of a free group has infinite order.

We begin with a lemma.

Lemma: Let $w \in F(S)$. Denote by $\mathsf{len}(w)$ the reduced length of $w$. Then $\mathsf{len}(w^2) > \mathsf{len}(w)$. Proof: We proceed by induction on $\mathsf{len}(w)$. If $w$ has length 1, then $w = a$ for some letter $a$. Then $w^2 = a^2$, and this word is reduced; hence $\mathsf{len}(w^2) = 2 > 1 = \mathsf{len}(w)$. Suppose $\mathsf{len}(w) = 2$. There are two cases; if $w = a^2$, then $\mathsf{len}(w^2) = 4 > 2 = \mathsf{len}(w)$. If $w = ab$, where $b \neq a$, then there are again two cases. If $b = a^{-1}$, then $w = 1$, a contradiction. If $b \neq a^{-1}$, then $w^2 = abab$ is reduced, and we again have $\mathsf{len}(w^2) > \mathsf{len}(w)$. Suppose now that every word of length at most $k$ satisfies $\mathsf{len}(w^2) > \mathsf{len}(w)$ for some $k \geq 2$. Let $w = aub$ be a word of length $k+1$, where $a$ and $b$ are letters and $u$ is a word. Now $w^2 = aubaub$. If $b = a^{-1}$, then $w^2 = au^2b$. Since $u$ is a reduced word of length at least 1, $\mathsf{len}(u^2) > \mathsf{len}(u)$. Now $\mathsf{len}(w^2) \geq \mathsf{len}(u^2) + 2$ $> \mathsf{len}(u) + 2$ $= \mathsf{len}(w)$. If $b \neq a^{-1}$, then $w^2 = aubaub$ is reduced, and we have $\mathsf{len}(w^2) = 2 \mathsf{len}(w) > \mathsf{len}(w)$. $\square$

Let $w \in F(S)$ be a reduced word with $w \neq 1$. Then the length of $w$ is positive, so that by the lemma, $\mathsf{len}(w^2) > \mathsf{len}(w)$. Moreover, the words $w^{2^k}$ have arbitrarily large reduced length. But if $|w|$ is finite, there exists a maximal reduced length among the reduced lengths of powers of $w$.

### A group of order 24 with no elements of order 6 is isomorphic to Sym(4)

Show that a group of order 24 with no element of order 6 is isomorphic to $S_4$.

Note that $24 = 2^3 \cdot 3$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) = 1$. If $n_2(G) = 1$, then by the recognition theorem for direct products, $G \cong P_2 \times P_3$, where $P_2$ and $P_3$ are Sylow 2- and 3-subgroups of $G$, respectively. By Cauchy, there exist elements $x \in P_2$ and $y \in P_3$ of order 2 and 3, so that $xy$ has order 6, a contradiction. Suppose now that $n_2(G) = 3$. Since $n_2(G) \not\equiv 1$ mod 4, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $P_2 \cap Q_2$ is nontrivial and its normalizer has order $2^3 \cdot 3$, by a previous theorem. Thus $P_2 \cap Q_2 \leq G$ is normal. Note that $|P_2 \cap Q_2|$ is either 2 or 4.

Suppose $|P_2 \cap Q_2| = 4$. Then at most $7 + 3 + 7$ nonidentity elements of $G$ are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that $|P_2 \cap Q_2| = 2$. By the N/C Theorem, $G/C_G(P_2 \cap Q_2) \leq \mathsf{Aut}(P_2 \cap Q_2) \cong 1$, so that $P_2 \cap Q_2$ is central in $G$. By Cauchy, there exist elements $x \in P_2 \cap Q_2$ and $y \in G$ of order 2 and 3, so that $xy$ has order 6, a contradiction.

Thus we may assume $n_3(G) = 4$. Let $P_3 \leq G$ be a Sylow 3-subgroup and let $N = N_G(P_3)$. The action of $G$ on $G/N$ yields a permutation representation $G \rightarrow S_4$ whose kernel $K$ is contained in $N$. Recall that normalizers of Sylow subgroups are self normalizing, so that $N$ is not normal in $G$. Moreover, we have $|N| = 6$. We know from the classification of groups of order 6 that $N$ is isomorphic to either $Z_6$ or $D_6$; however, in the first case we have an element of order 6, a contradiction. Thus $N \cong D_6$. We know also that the normal subgroups of $D_6$ have order 1, 3, or 6. If $|K| = 6$, then $K = N$ is normal in $G$, a contradiction. If $|K| = 3$, then by the N/C theorem we have $G/C_G(K) \leq \mathsf{Aut}(Z_3) \cong Z_2$. In particular, $C_G(K)$ contains an element of order 2, so that $G$ contains an element of order 6, a contradiction.

Thus $K = 1$, and in fact $G \leq S_4$. Since $|G| = |S_4| = 24$ is finite, $G \cong S_4$.

### A finite group is nilpotent if and only if all pairs of elements of relatively prime order commute

Prove that a finite group $G$ is nilpotent if and only if whenever $a, b \in G$ with $\mathsf{gcd}(|a|,|b|) = 1$, then $ab = ba$. [Hint: Use Theorem 3.]

$(\Rightarrow)$ Suppose $G$ is a finite nilpotent group. Let $a,b \in G$ with $\mathsf{gcd}(|a|,|b|) = 1$. By Theorem 6.3, $G$ is the internal direct product $G = P_1P_2 \cdots P_k$ of its Sylow subgroups; thus we may write $a = a_1a_1\cdots a_k$ and $b = b_1b_2 \cdots b_k$, where $a_i,b_i \in P_i$. Moreover, $|a| = \mathsf{lcm}(|a_i|)$ and $|b| = \mathsf{lcm}(|b_i|)$; if some $a_i \neq 1$, then $p_i$ divides $|a|$. so $p_i$ does not divide $|b|$, and $b_i = 1$. Similarly, if $b_i \neq 1$ then $a_i = 1$.

By relabeling the Sylow subgroups of $G$ and collecting factors, we have the internal direct product $G = Q_1 \times Q_2$, with $a \in Q_1$ and $b \in Q_2$. Thus $ab = ba$.

$(\Leftarrow)$ Suppose $G$ is a finite group in which the hypothesis holds. Let $P \leq G$ be a Sylow $p$-subgroup. If $Q \leq G$ is a Sylow subgroup for some other prime $q$, then $Q \leq C_G(P) \leq N_G(P)$. In particular, $N_G(P)$ contains a subgroup of maximal $q$-power order for each $q$ dividing $|G|$; thus $N_G(P) = G$, so that $P \leq G$ is normal. Since $P$ is arbitrary, all Sylow subgroups of $G$ are normal. By Theorem 3, $G$ is nilpotent.

### A criterion for the existence of unique group homomorphisms from an abelian group

Let $A = \langle x_1 \rangle \times \cdots \langle x_t \rangle$ be a finite abelian group with $|x_i| = n_i$ for each $i$.

1. Find a presentation for $A$.
2. Prove that if $G$ is a group containing commuting elements $g_1,\ldots,g_t$ such that $g_i^{n_i} = 1$ for each $i$, then there is a unique group homomorphism $\theta : A \rightarrow G$ such that $\theta(x_i) = g_i$.

1. We claim that $A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle$, where $1 \leq i,j \leq t$. Indeed, the “standard basis” elements $e_i$, consisting of $x_i$ in the $i$-th coordinate and 1 in all other coordinates, satisfy these relations.
2. Note that every element of $A$ can be written uniquely as $(x_i^{b_i})_{i=1}^t$. Define $\theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}$, where the product on the right hand side is as computed inside $G$. It is clear that $\theta(x_i) = g_i$ for each $i$. Moreover, $\theta$ is a homomorphism because the $g_i$ commute pairwise.

Finally, suppose $\sigma$ is a homomorphism $A \rightarrow G$ such that $\sigma(x_i) = g_i$ for each $i$. Then $\sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i}$ $= \prod \theta(x_i)^{b_i}$ $= \theta((x_i^{b_i}))$. Thus $\sigma = \theta$ (products computed inside $G$), so that $\theta$ is unique.

### A divisibility criterion for orders of elements in a finite abelian group

Let $G$ be a finite abelian group of type $(n_1,n_2,\ldots,n_k)$. (That is, $n_1,\ldots,n_k$ are the invariant factors of $G$.) Prove that $G$ contains an element of order $m$ if and only if $m|n_1$. Deduce that $G$ is of exponent $n_1$.

$(\Rightarrow)$ Suppose $G$ contains an element $x$ of order $m$. Now $x = (a_1,a_2,\ldots,a_k)$ for some $a_i \in Z_{n_i}$, and we have $m = \mathsf{lcm}(|a_1|,|a_2|,\ldots,|a_k|)$. By the divisibility criterion for invariant factors, $n_1$ is a common multiple of the $n_i$, hence of the $|a_i|$. Thus $m|n_1$.

$(\Leftarrow)$ Suppose $m|n_1$. Now $G$ has a subgroup isomorphic to $Z_{n_1}$, which (being cyclic) has an element of order $m$. Thus $G$ has an element of order $m$.

We conclude that if $x \in G$ has order $m$, then since $n_1 = tm$ for some $t$, $n_1x = tmx = 0$. Hence the exponent of $G$ is finite and divides $n_1$. Moreover, the exponent of $G$ is at least $n_1$ since $G$ has an element of order $n_1$. Thus $G$ has exponent $n_1$.

### Count the elements of order 7 in a simple group of order 168

How many elements of order 7 must there be in a simple group of order 168?

Note that $168 = 2^3 \cdot 3 \cdot 7$. If $G$ is a simple group of order 168, Sylow’s Theorem forces $n_7 = 8$. Moreover, the Sylow 7-subgroups intersect trivially, and every element of order 7 is in some Sylow 7-subgroup. Thus the number of elements of order 7 is $8 \cdot 6 = 48$.

### Compute the number of conjugacy classes of elements of prime order in Sym(n)

Let $p$ be a prime. Find a formula for the number of conjugacy classes of elements of order $p$ in $S_n$ using the floor function.

Every element of order $p$ in $S_n$ is a product of commuting $p$-cycles. Provided $mp \leq n$, there is a conjugacy class in $S_n$ consisting of all products of $m$ commuting $p$-cycles; that is, one class for all integers $1\leq m \leq n/p$. Thus the number of such classes is $\lfloor n/p \rfloor$.

### Every element of order 2 in Alt(n) is the square of an element of order 4 in Sym(n)

Prove that every element of order 2 in $A_n$ is the square of an element of order 4 in $S_n$.

We know that every element of order 2 in $S_n$ (hence also $A_n$) is a product of commuting 2-cycles by a previous theorem. Write $\sigma \in A_n$ of order 2 as $\sigma = (a_{1,1}\ a_{1,2})(b_{1,1}\ b_{1,2})(a_{2,1}\ a_{2,2})(b_{2,1}\ b_{2,2}) \cdots (a_{2k,1}\ a_{2k,2})(b_{2k,1}\ b_{2k,2})$. Note that the number of 2-cycles in the decomposition of $\sigma$ is even since $\sigma \in A_n$.

Note that $(a_{i,1}\ b_{i,1}\ a_{i,2}\ b_{i,2})^2 = (a_{i,1}\ a_{i,2})(b_{i,1}\ b_{i,2})$. Thus $\sigma = (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})^2 (a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2})^2 \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2})^2$ $= \left( (a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2}) \right)^2$ using a previous theorem, and the element $(a_{1,1}\ b_{1,1}\ a_{1,2}\ b_{1,2})(a_{2,1}\ b_{2,1}\ a_{2,2}\ b_{2,2}) \cdots (a_{2k,1}\ b_{2k,1}\ a_{2k,2}\ b_{2k,2})$ has order 4 in $S_n$ by §1.3 #15.