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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 $G$)?

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    $\begingroup$ 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 $G=H^n$.... $\endgroup$ – Greg Martin Aug 13 at 5:34
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Aug 13 at 5:40
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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.

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