If |G/Z(G)| = n, then G is n-abelian

Posted: April 15, 2024 in Basic Algebra, Groups
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As we defined here, given an integer n, we say that a group G is n-abelian if (xy)^n=x^ny^n for all x,y \in G. Here we showed that if G/Z(G), where Z(G) is the center of G, is abelian and if |Z(G)|=n is odd, then G is n-abelian. We now prove a much more interesting result.

Proposition. Let G be a group with the center Z(G). If |G/Z(G)|=n, then G is n-aberlian.

Proof. Let \{g_1, g_2, \cdots , g_n\} be a transversal of Z(G) in G, as defined in this post. Let x \in G, and let \alpha \in S_n, h_i \in Z(G), \ 1 \le i \le n, be such that xg_i=g_{\alpha(i)}h_i. Then

x=g_{\alpha(i)}g_i^{-1}h_i. \ \ \ \ \ \ \ \ \ \ \ \ (1)

Let (i \ \alpha(i) \ \alpha^2(i) \ \cdots \ \alpha^{k-1}(i)) be any cycle in the decomposition of \alpha into disjoint cycles. Then, by (1),

x^k=g_{\alpha(i)}g_i^{-1}g_{\alpha^k(i)}g_{\alpha^{k-1}(i)}^{-1}g_{\alpha^{k-1}(i)}g_{\alpha^{k-2}(i)}^{-1} \cdots g_{\alpha^2(i)}g_{\alpha(i)}^{-1}h_ih_{\alpha(i)} \cdots h_{\alpha^{k-1}(i)}

and so since \alpha^k(i)=i, we get that

x^k=h_ih_{\alpha(i)} \cdots h_{\alpha^{k-1}(i)}. \ \ \ \ \ \ \ \ \ \ \ \ (2)

If \alpha is the product of m disjoint cycles, then we will have m identities of the form (2). Multiplying all the m identities together gives x^n=h_1h_2 \cdots h_n. On the other hand, by the Theorem in the post linked at the beginning of the proof, the map \lambda : G \to Z(G) defined by \lambda(x)=h_1h_2 \cdots h_n is a group homomorphism. Hence \lambda(x)=x^n is a group homomorphism and so, for all x,y \in G,

x^ny^n=\lambda(x)\lambda(y)=\lambda(xy)=(xy)^n,

proving that G is n-abelian. \ \Box

Note. The Proposition is Lemma 6.22 in T. Y. Lam’s book A First Course in Noncommutative Rings. However, Lam does not give a proof of \lambda(x)=x^n; the proof was added by me.

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