Why does $G$ finite abelian group with $i_G$ involutions have no subgroups isomorphic to $K_4$ if $6\nmid i_G(i_G-1)$?

Question

I was reviewing my answer here, and have realized that the provided solution ought to work for any finite abelian group. The generalization would then be:

Claim. Let $G$ be a finite abelian group and $i_G$ the number of elements of order $2$ of $G$. Then, the number of subgroups of $G$ isomorphic to Klein's $4$-group, say $n_K$, is given by:

$$n_K=\frac{1}{6}\cdot i_G(i_G-1)$$

As a corollary, this would mean that, if a finite abelian group $G$ has $i_G$ elements of order $2$ such that $6\nmid i_G(i_G-1)$, then $G$ has no subgroups isomorphic to $K_4$.

Provided that this not based on false assumptions (the linked answer and/or its generalization), what's the group-theoretic motivation for that?

Answer 1

Let $G$ be an abelian group. Then we have $G=G_2\times G_{2'}$, where $G_2$ is the Sylow $2$-subgroup of $G$ and $G_{2'}$ is the Hall $2'$-subgroup of $G$.

It suffices to consider $G_2$ if we do with involutions. Note also that \begin{equation*} G_2=\mathbb{Z}_{2^{e_1}}\times\cdots\times\mathbb{Z}_{2^{e_k}} \end{equation*} for some $1\le e_1\le\cdots\le e_k$, and each $\mathbb{Z}_{2^{e_i}}$ gives exactly one involution. Therefore, $i_G=i_{G_2}=2^k-1$.

If $k=0$ then $G$ is odd. In this case of course $G$ has no involution.

If $k=1$ then it is also straightforward to see that $n_K=0$.

If $k\ge 2$ then any two distinct involutions generate a $K_4$.

Your claim is correct. But $6\nmid i_G(i_G-1)$ if and only if $2^k-1=i_G\equiv 2\pmod 3$, which can never happen.

Answer 2

The number of involutions is $2^k-1$ for some $k$, and $6$ always divides $(2^k -1)(2^k-2)$.

Answer 3

Indeed, if $a,b$ are distinct elements of order $2$, then $0,a,b,a+b$ form a Klein 4-group. Each such subgroup is obtained in exactly $3$ ways, whence the claim. It follows that there is a Klein 4 as soon as $i_G\ge 2$.