Conjugates Generating a normal subgroup

Question

This problem is taken from I.N. Herstein

Problem:
If $G$ is a group and $a \in G$ if of finite order and has only a finite number of conjugates in $G$, prove that these conjugates of $a$ generate a finite normal subgroup of $G$.

I would like to see a solution for this and also I would like to know whether the conjugates always generate a normal subgroup.

@The Community: Please see this post on meta regarding the phrasing of questions and discuss (over there). — Tom Stephens, Aug 15 '10 at 19:24
@Robin Chapman: Yes sir, its from "Topics in Algebra" second edition, supplementary problem section at the end of chapter two. — , Aug 15 '10 at 19:24
The subgroup generated by an element of a group and its conjugates is always normal. — Robin Chapman, Aug 15 '10 at 19:38
It is also one of the background results used in http://math.stackexchange.com/questions/2355/is-there-a-group-with-exactly-92-elements-of-order-3 — Jack Schmidt, Aug 15 '10 at 23:35
@Robin: I think the content of the question is that the group generated by the finitely many conjugates is still finite (no-one said $G$ itself was finite). — Geoff Robinson, Jul 10 '11 at 09:26

score 3 · Accepted Answer · answered Aug 15 '10 at 23:34

3

This called Dicman's lemma and is 14.5.7 on page 425 (1st ed) or page 442 (2nd ed) of Robinson's Course in the Theory of Groups.

answered Aug 15 '10 at 23:34

Jack Schmidt

55,589

It looks to me like this is the special case of Dicman's Lemma in which the finite normal subset consists of conjugates of a single element. Perhaps this is a little easier than the general case? (Although the reason I write this now is that I just put the proof of Dicman's Lemma (from Robinsons's text) into some notes of mine. The proof is almost as easy as it could possibly be...) – Pete L. Clark Jan 22 '17 at 22:59

Conjugates Generating a normal subgroup

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