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Let G be a finite non abelian group without element of order 2, and denote commutator subgroup of G as G', normal subgroup of G as H, G' $\leq$ H $\triangleleft$ G. Show the product of all elements in G is in the normal subgroup H.

I attempted to use correspondence theorem but it seems to be wrong, below is what I tried:

denote elements in G/G' as $\tilde{g}$ , then $g_1 g_2... \in \tilde{g_1}G'\tilde{g_2}G' = h_1G'G'h_2G'G'... \in H$

Any help is really appreciated!

Hao-Sen Wu
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  • If $G'\le H$, then we can show that $G/H$ is Abelian. What do you exactly want? Can't understand your question. – QED Dec 13 '17 at 05:56
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    The answer in the link https://math.stackexchange.com/questions/1918438/product-of-all-elements-in-finite-nonabelian-group?noredirect=1&lq=1 may be helpful. – Phil. Z Dec 13 '17 at 05:57
  • thanks but this question is quite different from mine – Hao-Sen Wu Dec 13 '17 at 06:17
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    @Hao-SenWu The answer in the link said that the set of all different products of all the elements of $G$ is exactly the commutator subgroup $G'$, provided the $2$-sylow subgroup of $G$ is trivial. Does that answer your question? Or maybe I missed some points? – Phil. Z Dec 13 '17 at 06:37

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A key fact here is that if $a,b \in G$ a group, then $ab = ba[a,b]$.

And so if $|G|$ is odd, within an expression of the product of all elements of $G$, one can switch two adjacent elements at the "expense" of a commutator element. And so after moving each non-identity element $g$ to be adjacent to the different element, $g^{-1}$, this product of all of the elements of $G$ reduces to a product of commutator elements of $G$.

He Is Alive gamme
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