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Let G be order of 8,if G has a element of order 2 not lying center,how to prove G is isomorphic to $D_8$?

A hint is consider the sylow-2 subgroup of $S_4$, I know it's $D_8$. So I want to construct a group action to let $G$ embed in $S_4$, but I can't find the action.

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2 Answers 2

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Let $g\in G$ be the element of order two not lying in the center. Let $G$ act by conjugation on the set of left cosets of $G/\langle g\rangle .$

So you get a homomorphism from $G$ into $S_4.$

The kernel is $\bigcap_{h\in G}h\langle g\rangle h^{-1}=\{e\},$ because $g$ isn't in the center (prove!).

Thus $G$ is (isomorphic to) the Sylow-$2$ subgroup of $S_4.$

Btw, this is one of the most indispensable actions to know about.

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    $\begingroup$ The part of ker is the key! Thank you for your answer! $\endgroup$
    – ckx
    Apr 11 at 6:05
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    $\begingroup$ The kernel of that action is called the normal core of the subgroup you quotient by. $\endgroup$ Apr 11 at 6:07
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    $\begingroup$ I think it should action on all conjugate class of G about H but not left closet. $\endgroup$
    – ckx
    Apr 11 at 6:14
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Off-hint. Firstly, of course, such a $G$ is nonabelian as there is a noncentral element. Secondly, the existence of a centerless $G$ of order $8$ is prevented by the class equation and the fact the $7$ can't be summed up with terms which are powers of $2$. Moreover, $|Z(G)|\ne 4$, as there isn't any group with nontrivial cyclic quotient $G/Z(G)$. Therefore, $|Z(G)|=2$, and hence, by assumption, there are at least two elements of order $2$. If you know that the quaternion group has one element of order $2$, only, then you are left with $G$ being $D_8$ ($D_4$ for many).

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    $\begingroup$ Fine, but your solution relies on the classification of non-abelian groups of order $8$, which is not assumed … $\endgroup$ Apr 11 at 14:30
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    $\begingroup$ Agreed, @NickyHekster. $\endgroup$
    – Kan't
    Apr 11 at 16:52

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