Return to Answer

This is a theorem of Bernhard Neumann and his wife (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

Edit 1: The paper is this: MR0064042 Neumann, B. H.; Neumann, Hanna "Partial endomorphisms of finite groups." J. London Math. Soc. 29, (1954). 434–440.

If $G$ and $H$ are given as subgroups in the same finite group $F$, then there is no need to take $F= G\times H$.

Edit 2: In fact the paper by Neumanns cited above proves a much stronger result when there is only a homomorphism from $G$ to $H$ (the question is then whether it extends to an endomorphism of a bigger finite group). The fact about arbitrary isomorphic finite subgroups is proved first in Neumann, B. H. (notice the absence of Hanna Neumann) An essay on free products of groups with amalgamations. Philos. Trans. Roy. Soc. London. Ser. A. 246, (1954). 503–554. The paper gives two proofs of this fact (Corollary 18.3). The first uses amalgamated products and is essentially the proof above. The second proof is attributed to Ph. Hall and is essentially the same as Will Savin's proof discussed in the comments.

This is a theorem of Bernhard Neumann and his wife (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

Edit 1: The paper is this: MR0064042 Neumann, B. H.; Neumann, Hanna "Partial endomorphisms of finite groups." J. London Math. Soc. 29, (1954). 434–440.

If $G$ and $H$ are given as subgroups in the same finite group $F$, then there is no need to take $F= G\times H$.

Edit 2: In fact the paper by Neumanns cited above proves a much stronger result when there is only a homomorphism from $G$ to $H$ (the question is then whether it extends to an endomorphism of a bigger finite group). The fact about arbitrary isomorphic finite subgroups is proved first in Neumann, B. H. (notice the absence of Hanna Neumann) An essay on free products of groups with amalgamations. Philos. Trans. Roy. Soc. London. Ser. A. 246, (1954). 503–554. The paper gives two proofs of this fact (Corollary 18.3). The first uses amalgamated products and is essentially the proof above. The second proof is attributed to Ph. Hall and is essentially the same as Will Sawin's proof discussed in the comments.

This is a theorem of Bernhard Neumann and his wife (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

Edit: The paper is this: MR0064042 Neumann, B. H.; Neumann, Hanna "Partial endomorphisms of finite groups." J. London Math. Soc. 29, (1954). 434–440.

If $G$ and $H$ are given as subgroups in the same finite group $F$, then there is no need to take $F= G\times H$.

This is a theorem of Bernhard Neumann and his wife (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

Edit 1: The paper is this: MR0064042 Neumann, B. H.; Neumann, Hanna "Partial endomorphisms of finite groups." J. London Math. Soc. 29, (1954). 434–440.

If $G$ and $H$ are given as subgroups in the same finite group $F$, then there is no need to take $F= G\times H$.

Edit 2: In fact the paper by Neumanns cited above proves a much stronger result when there is only a homomorphism from $G$ to $H$ (the question is then whether it extends to an endomorphism of a bigger finite group). The fact about arbitrary isomorphic finite subgroups is proved first in Neumann, B. H. (notice the absence of Hanna Neumann) An essay on free products of groups with amalgamations. Philos. Trans. Roy. Soc. London. Ser. A. 246, (1954). 503–554. The paper gives two proofs of this fact (Corollary 18.3). The first uses amalgamated products and is essentially the proof above. The second proof is attributed to Ph. Hall and is essentially the same as Will Savin's proof discussed in the comments.

This is a theorem of Bernhard Neumann (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

This is a theorem of Bernhard Neumann and his wife (and, as he once told me, can be found in his Collection of Works; he even told me which volume I should look at, but I do not remember; I am sure mathsci can help). A "modern" proof of that fact is easy. Basically suppose that two finite groups $G$ and $H$ are isomorphic. Then take $F=G\times H$ and the HNN extension $T$ of $F$ with associated subgroups $G, H$. In $T$, the subgroups $G$ and $H$ are conjugate. Now, $T$ is infinite, but as every HNN extension of a finite group, $T$ is residually finite because it is virtually free (say, by Stallings ends theorem). Hence $T$ has a finite quotient $R$ such that the natural maps from $G$ and $H$ into $R$ are embeddings. Thus $G$ and $H$ are conjugated subgroups of a finite group $R$.

Edit: The paper is this: MR0064042 Neumann, B. H.; Neumann, Hanna "Partial endomorphisms of finite groups." J. London Math. Soc. 29, (1954). 434–440.

If $G$ and $H$ are given as subgroups in the same finite group $F$, then there is no need to take $F= G\times H$.