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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.

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    Is this from a book by Herstein? If so which one?2010-08-15
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    @The Community: Please see [this](http://meta.math.stackexchange.com/questions/588/how-can-i-ask-a-good-question/647#647) post on meta regarding the phrasing of questions and discuss (over there).2010-08-15
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    @Robin Chapman: Yes sir, its from "Topics in Algebra" second edition, supplementary problem section at the end of chapter two.2010-08-15
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    The subgroup generated by an element of a group and its conjugates is always normal.2010-08-15
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    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-32010-08-15
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    @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).2011-07-10

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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.

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    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...)2017-01-22