I'm working on a problem but it's a bit complicated to express. Here's the setup: $(G,\cdot)$ and $(H,\ast)$ are groups such that (crucially) $G\cap H = \emptyset$. Moreover, the operations $\cdot$ and $\ast$ extend, respectively, to group operations $\circ$ and $\diamond$ on $G\cup H$ that are compatible with respect to parentheses. That is, $G\leq (G\cup H, \circ)$ and $H\leq (G\cup H, \diamond)$, and for any $a,b,c\in G\cup H$, we have $(a\circ b)\diamond c = a\circ (b\diamond c)$ and $(a\diamond b)\circ c = a\diamond (b\circ c)$.
In this situation, it's clear that the identity $e_G$ of $(G,\cdot)$ serves as the identity for $(G\cup H,\circ)$, while $e_H\in H$ is the identity with respect to $\diamond$.
My question is as follows:
Must $e_H$ have order 2 with respect to the operation $\circ$, i.e., does $e_H\circ e_H = e_G?$ (Or, equivalently, does $e_G\diamond e_G = e_H$?)
Why in the world should I expect this? Well, in the one example of this kind of $(\circ,\diamond)$-system that I can construct (please see my other recent post: Necessary and sufficient condition for a pair of groups to be isomorphic) these conditions do hold. Also, I can see that the order of $e_H$ in the group $(G\cup H, \circ)$, if finite, must be even, since multiplying an element of $H$ by itself an odd number of times in the operation $\circ$ yields an element of $H$ (but the identity of the group $(G\cup H, \circ)$ lies in $G$). But I can't seem to deduce that the order is just 2, at least not without tacking on additional assumptions.
My other motivation is this: After much tedious "bookkeeping," I was able to prove the following correspondence theorem:
Theorem: Let $(G,\cdot)$ and $(H,\ast)$ be groups such that $G\cap H=\emptyset$. Then the (possibly empty) set of all group isomorphisms $f:G\to H$ is in one-to-one correspondence with the set of all algebraic systems $(G\cup H,\circ, \diamond)$ satisfying the following properties:
$\circ$ and $\diamond$ are compatible (in the above sense) group operations on $G\cup H$
$\cdot = \circ|_{G\times G}$ and $\ast = \diamond|_{H\times H}$
The identity $e_H$ of the group $(H,\ast)$ has order 2 with respect to the operation $\circ$, and belongs to the center of the group $(G\cup H,\circ)$.
The proof relies on the observation that any isomorphism $f:G\to H$ induces a pair of compatible group operations $\circ_f$, $\circ_{f^{-1}}$ on the union that extend the operations on $G$ and $H$, as discussed in the linked post. The map
\begin{equation} f\mapsto (G\cup H,\circ_f, \circ_{f^{-1}}) \end{equation}
is the desired bijection, although this isn't, as far as I can tell, remotely obvious. Basically, the function $f(g)=g\circ e_H$ ends up being an isomorphism $G\to H$, and condition (3) allows us to prove that $\circ=\circ_f$ and $\diamond = \circ_{f^{-1}}$ for this $f$, establishing surjectivity. I'm happy to elaborate if need be, but I'd like to avoid getting side-tracked here.
Now, the statement of this theorem would, I feel, be much more elegant if one or both of the assumptions in (3) ended up being superfluous, i.e., provable from (1) and (2). So, I'm hoping that I can at least show that the order 2 condition always holds. But I haven't been able to prove this. I tried constructing a counter-example using groups of small order, but got a bit lost in checking the compatibility conditions.
Here are a few other observations that I've made: The inverse of $g\in G$ with respect to the operation $\diamond$ is the element $e_H\circ g^{-1}\circ e_H$, where $g^{-1}\in G$ is the original inverse of $g$. A dual formula holds for the inverse of $h\in H$ with respect to $\circ.$ Also, if $G$ and $H$ are finite, then since the order of the disjoint union is 2|G|, Cauchy's Theorem implies that there must be an element of order 2 in $(G\cup H,\circ).$ If $|G|$ happens to be odd, then this order 2 element must belong to the set $H$, by Lagrange's Theorem. But is it $e_H$?
Any suggestions or insights into this problem would be greatly appreciated. I'm sorry for the long and involved question, but hopefully it's interesting and I'm not missing something simple.
Thank you.
Let $\Gamma = \mathbf Z/4\mathbf Z$. Define $\circ,\diamond:\Gamma^2\to\Gamma$ by $x \circ y = x + y$ and $x \diamond y = x + y + 1$. The associativity rules hold. Let $G = (\{0,2\}, \circ)$ and $H = (\{1,3\}, \diamond)$. Here $e_G = 0$ and $e_H = 3$, and $0\diamond0 = 1 \ne 3$.