Prove Abelian groups with two elements of order 2 have a subgroup of order 4.

Question

Prove Abelian groups with two elements of order 2 have a subgroup of order 4. (From Gallian Algebra).

My proof is below, as an answer.

1. Can you verify, critique, or improve my proof, or the proof writing; or provide an alternate or simpler proof?

2. I proved this "manually," via construction. Is there a simpler, more direct, or more abstract approach?

3. My proof is limited to the exact question asked. Is there a more universal approach that would work in similar questions as well?

Answer 1

Your proof is fine.

More generally, if $G$ is an Abelian group with $H$ and $K$ subgroups thereof, the set $$ HK=\{hk:h\in H,k\in K\} $$ is a subgroup of $G$ (this holds also in general groups, provided one of the subgroups is normal).

In the case of Abelian groups, though, there is a further interesting property: the map $$ \mu\colon H\times K\to HK,\qquad \mu(h,k)=hk $$ is a group homomorphism (easy verification); the domain is the product group. What's the kernel? We have $\mu(h,k)=e$ if and only if $k=h^{-1}$, hence the kernel is $$ \ker\mu=\{(x,x^{-1}):x\in H\cap K\} $$ By the homomorphism theorem, we know that, in case of finite groups, $$ |HK|=\frac{|H\times K|}{\lvert\ker\mu\rvert}=\dfrac{|H|\,|K|}{|H\cap K|} $$ Note that $H\cap K$ is not $\ker\mu$, but they have the same order.

Now specialize this to $H=\langle a\rangle$ and $K=\langle b\rangle$, where $a,b\in G$ have order $2$ and $a\ne b$.

Then $\langle a\rangle\cap \langle b\rangle=\{e\}$, because it cannot contain $a$ and is a subgroup of $\langle a\rangle$. Hence $$ |\langle a\rangle\langle b\rangle|=\frac{2\cdot2}{1}=4 $$ Since $e,a,b,ab\in\langle a\rangle\langle b\rangle$, this is the full list of elements.

Can we generalize it? Yes, try the case of two elements of order the prime $p$; however, $a\ne b$ is no longer sufficient and something more has to be required, precisely that neither is a power of the other.

Answer 2

Let $G$ be an Abelian group with elements $a,b$ such that $|a| = |b| = 2, a \neq b$.

The subgroup $H = {e, a, b, ab}$ is of order 4. $ab \neq a$ or $b$, since neither $a$ or $b$ is the identity. $ab \neq e$ since $aa = e, a \neq b$, and an element can have only one inverse.

$H$ includes all its inverses, as $(ab)^{-1} = ab$, since $abab=aabb=e$.

$H$ is closed, as $aba = aab = b$ and $bab = bba = b$.