# G has a element of order 2 not lying in center

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.

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.

• The part of ker is the key! Thank you for your answer!
– ckx
Apr 11 at 6:05
• The kernel of that action is called the normal core of the subgroup you quotient by. Apr 11 at 6:07
• I think it should action on all conjugate class of G about H but not left closet.
– ckx
Apr 11 at 6:14

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).

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