Group in which every nonidentity element is of order 2

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!

• Why the downvote? There's ample context! Oct 14, 2019 at 17:03

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 $$V$$.
• 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 $H$? Oct 14, 2019 at 19:48