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.