# Extension of Group Isomorphism

Suppose $H_{1},H_{2}$ are finite subgroups of a finitely generated abelian group, $G$. Suppose that $\phi$ is an isomorphism from $H_{1}$ to $H_{2}$. Can $\phi$ be extended to an automorphism on $G$?

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In $G = \mathbb Z/2\mathbb Z \times \mathbb Z/4\mathbb Z$ the unique isomorphism between the groups of order $2$ generated by $\langle 0,2 \rangle$ and $\langle 1,0 \rangle$ can’t be extended to an automorphism of $G$; where would $\langle 0,1 \rangle$ go?

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In general, no (as examples already given show). However, if you modify the situtation a little bit, you can get even more; specifically:

Let $G$ be a group, and let $H_1$ and $H_2$ be two subgroups of $G$. If $\varphi\colon H_1\to H_2$ is an isomorphism, then there exists an overgroup $K$ of $G$, and an element $t\in K$, such that for all $h\in H_1$, $t^{-1}ht=\varphi(h)$. That is, the isomorphism $\varphi$ extends to an inner automorphism in an overgroup of $G$.

This is handled by the construction of HNN Extensions, named after Graham Higman, Bernhard Neumann, and Hanna Neumann. The group can be realized as follows: if $G=\langle X\mid R\rangle$ is a presentation of $G$, let $t$ be a symbol not in $X$. Then the $HNN$-extension with base $G$ extending $\varphi$ is $$G_{*\varphi} = \langle X,\; t\>\mid\> S, T\rangle$$ where $$T = \{ t^{-1}ht\varphi(h)^{-1}\mid h\in H_1\}.$$

Note, however, that $K$ is necessarily not abelian.

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