Subgroups of $G\times G$ that are isomorphic with $G$

Let $$G$$ be a finite group. I want to find all subgroups $$H$$ of $$G\times G$$ such that $$H\cong G$$.

It's easy to find three such subgroups:

(1) $$\{(g,1):g\in G\}$$,

(2) $$\{(1,g):g\in G\}$$ and

(3) the diagonal $$\Delta(G) = \{(g,g): g\in G\}$$.

More generally, for any $$\sigma \in \mathrm{End}(G)$$, $$\{(g,g^\sigma):g\in G\}$$ and $$\{(g^\sigma,g):g\in G\}$$ are also subgroups of $$G$$ that are isomorphic with $$G$$. Therefore, we have $$2|\mathrm{End}(G)|-1$$ such subgroups.

If $$G$$ is decomposable, say $$G = H\times K$$, then $$H_1\times K_2\cong G$$ is also a subgroup of $$G_1\times G_2 = (H_1\times K_1)\times (H_2\times K_2)$$, where $$G_1,G_2\cong G$$, $$H_1,H_2\cong H$$ and $$K_1,K_2\cong K$$. Hence we have more such subgroups.

My questions:

Are there more such subgroups that are not of types above? And what is the total number of such subgroups of $$G\times G$$ (given $$G$$)?

• I think you can generalize both your families of examples to $\{(g^{\sigma_1},g^{\sigma_2})\colon g\In G\}$ where $\sigma_1,\sigma_2$ are endomorphisms of $G$ such that ker$(\sigma_1)\cap{}$ker$(\sigma_2)=\{e\}$. There can be others: a cheap example is to take $G=H\times H$, so that $H\times\{e\}\times H\times\{e\}$ is a subgroup of $G\times G$ that technically isn't of the second form. And then imagine $G=H^n$.... – Greg Martin Aug 13 at 5:34
• Your count is not quite correct, since given an automorphism $\sigma$ of $G$ (which is an element of the endomorphism semigroup) you are counting both $\{(g^{\sigma},g)\mid g\in G\}$ and $\{(g,g^{\sigma^{-1}})\mid g\in G\}$... even though they are the same subgroup. – Arturo Magidin Aug 13 at 5:40

Your list includes the "anti-diagonal" $$\{(g,g^{-1}):g\in G\}$$ when $$G$$ is abelian. More generally there is $$\{(\alpha(g),\beta(g)):g\in G\}$$ where $$\alpha,\beta$$ are endomorphisms with trivially intersecting kernels.
In general, the subgroups of a group $$A$$ isomorphic to a group $$B$$ correspond to embeddings $$B\to A$$ mod the regular action of $$\mathrm{Aut}(B)$$. If $$A=G\times G$$ and $$B=G$$, any homomorphism $$G\to G\times G$$ is comprised of two homomorphisms $$\alpha,\beta:G\to G$$, and to be $$1$$-$$1$$ the kernels must intersect trivially.
Let $$n_H(G)$$ denote the number of subgroups of $$G$$ isomorphic to $$H$$. The number of embeddings $$H\to G$$ is then equal to $$e_H(G):=|\mathrm{Aut}(H)|n_H(G)$$ since the action of $$\mathrm{Aut}(H)$$ is regular. Then
$$e_G(G\times G)=\sum_{\substack{N,K\triangleleft G \\ N\cap K=1}} e_{G/N}(G)e_{G/K}(G).$$
Thus, we may compute $$n_G(G\times G)$$ by finding normal subgroups of $$G$$, finding how many subgroups of $$G$$ are isomorphic to the corresponding factor groups, finding the size of the automorphism groups of the factor groups, and finding which pairs of normal subgroups intersect trivially.