Commutator of normal subgroups is finite

Asked Sep 24 '17 at 14:44

Active Sep 24 '17 at 14:58

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$A, B \triangleleft G. X = \{[a, b]|a \in A, b \in B\}$. Prove that if $|X|< \infty $, then $|[A, B]|<\infty$.

By definition $[A, B] = \langle[a, b] | a \in A, b \in B\rangle$. So we must prove that by multiplication of these finite number of elements we can't get infinite number of elements. How can we do it?

edited Sep 24 '17 at 14:58

asked Sep 24 '17 at 14:44

Khan

Question title and body do not match. – Kenny Lau Sep 24 '17 at 14:45
Do not use generator symbol arbitrarily. – Kenny Lau Sep 24 '17 at 14:46
@KennyLau what's the problem? As I know $[A, B]$ is called commutator of subgroups – Khan Sep 24 '17 at 14:49
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You might like to look at https://math.stackexchange.com/questions/285852 which is a special case of your question. But if you want help then you say how far you have got and where you are stuck. – Derek Holt Sep 24 '17 at 17:19

Commutator of normal subgroups is finite

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