# Mutual Subgroups

Two groups can be proper subgroups of each other. Of course, they can’t be *finite* groups, but that’s alright.

A good place to start is with the free group on $n$ generators, which I’ll call $F^n$ (nonstandard notation, but fine when $n$ is finite). The free group $F^1$ is isomorphic to the integers under addition, so it isn’t very interesting. Let’s look at $F^2 = \langle x,y\rangle$ and $F^4 = \langle a,b,c,d\rangle$. It is obvious that $F^2 < F^4$. Of course there are multiple such inclusions, but the 12 most natural ones are $(x,y)\mapsto(a,b)$ and the like.

Now let’s construct an injection from $F^4$ to $F^2$, demonstrating that $F^4 < F^2$. A tempting guess is
\begin{equation*}
\phi : \left\{\begin{matrix}
a \mapsto xx\\

b \mapsto xy\\

c \mapsto yy\\

d \mapsto yx\\

\end{matrix}\right.\text.
\end{equation*}
Alas! This is indeed a group homomorphism, but it is not injective: you can verify that $\phi(b c^{-1} d a^{-1}) = 1$. A more robust encoding scheme (again, far from unique) is given by
\begin{equation*}
\phi : \left\{\begin{matrix}
a \mapsto x^0yx^{-0}\\

b \mapsto x^1yx^{-1}\\

c \mapsto x^2yx^{-2}\\

d \mapsto x^3yx^{-3}\\

\end{matrix}\right.\text.
\end{equation*}
(To see that this is injective, just think about how you would write a program to “decode” the string in $F^2$.)

Are $F^2$ and $F^4$ isomorphic? No. I like the proof that comes from looking at $\mathrm{Hom}(F^n, \mathbb Z / \mathbb Z_2)$ — that is, the set of all group homomorphisms from $F^n$ to the group with two elements. There are of course $2^n$ such homomorphisms, and that suffices to show that the structure of $F^n$ depends on $n$.

In conclusion: $F^2 < F^4$, $F^4 < F^2$, but $F^2 \not\approx F^4$.

It should be clear that none of this really depends on which free groups were chosen, as long as neither is $F^1$. In fact, we also have the glorious result $F^2 < F^2$ (and the same for any $F^n$ with $n \ge 2$). This is demonstrated by $(x,y)\mapsto(xx,yy)$. ($F^1 < F^1$ is easy to show as well.)