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I'm working on the following problem in Algebra:

Let $G$ be a group in which every nonidentity element is of order $2$. Show that every subgroup $H$ of $G$ has the property that $G/H$ is isomorphic to a subgroup of $G$.

Here's my progress so far:

First, I've shown that any group $G$ such that every nonidentity element is of order $2$ is abelian. That part is easy. Then, this means that every subgroup $H$ of $G$ is then normal, as every subgroup of an abelian group is normal ($\forall$ x $\in$ G & $\forall$ $h \in H$, $xhx^{-1} = xx^{-1}h = h \in H$ ).

Now, we recall that if $\phi:G \longrightarrow H$ is a group homomorphism, then $G/\ker(\phi) \cong \phi(G)$, where $\ker(\phi)$ is normal in $G$ by the First Isomorphism Theorem. Since every subgroup $H$ of $G$ is normal, & every normal subgroup is the kernel of a group homomorphism $\phi: G \longrightarrow G/H$, $G/H \cong \phi(G)$.

It's left to show that $\phi(G)$, the image of $\phi$, is isomorphic to a subgroup of $G$, where $\phi:G \longrightarrow G/H$ is a homomorphism for a normal subgroup $H$ of $G$. This is the last piece of the proof that I'm stuck on. Is my logic up to this point sound? If so, how can I show this last piece of the proof?

Thanks!

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    $\begingroup$ Why the downvote? There's ample context! $\endgroup$
    – Shaun
    Oct 14, 2019 at 17:03

1 Answer 1

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Such a group is abelian : if $(ab)^2=1$, then $ab ab=1$ but $a=a^{-1}, b=b^{-1}$, hence $aba^{-1}b^{-1}=1$, or$ab=ba$.

SO we can note the multiplication by a plus sign $(a+b)$ instead of $ab$, and $2.a=0$ for all $a$.

With this notation one sees that the group is in fact a $Z/2Z$ vector space, therefore isomorphic to $(Z/2Z)^d$, and a subgroup is isomorphic to $(Z/2Z)^e$.

I assumed that the dimension is finite, but using the existence of bases one can prove the same for infinite dimensional vector spaces. Namely if $V$ is a vector space and $W$ a subspace choosing a base of $V/W$ enable one to prove that it is isomorphic tp a subspace of $V$.

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  • $\begingroup$ Hi Thomas. I believe you're using the fact that if every nonidentity element of $G$ is of order $2$. then $G$ is isomorphic to a product of $Z/2Z$'s correct? And from here, you conclude that our subgroup $H$ of $G$ must then also be a product of $Z/2Z$'s? Thus, the quotient $G/H$ is again a product of $Z/2Z$'s? My only question is, how do we know that the number of $Z/2Z$'s showing up in the product corresponding to $G/H$ is equal to the number of $Z/2Z$'s showing up in the product corresponding to $H$? $\endgroup$ Oct 14, 2019 at 19:48
  • $\begingroup$ I transformed a question into a linear algebra problem, and the answer is : a quotient of a vector space is isomorphic to a subspace. $\endgroup$
    – Thomas
    Oct 15, 2019 at 20:10

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