Let $G$ be some finite group, $H_1, H_2 \leq G$ and $\phi:H_1 \mapsto H_2$ be a group isomorphism. Is it always possible to expand $\phi$ to an automorphism on $G$?

Explicitly, is there an automorphism $\sigma \in Aut_G$ s.t $\sigma \vert _{H_1}=\phi$?

Efforts so far

At first I considered defining $\sigma$ by letting $$\sigma\left(g\right)=\begin{cases} \phi\left(g\right) & g\in H_{1}\\ \phi^{-1}\left(g\right) & g\in H_{2}\\ g & else \end{cases}$$ but obviously such a $\sigma$ is not well-defined for elements in $H_1 \cap H_2$ (other than $e$). I then tried $$\sigma\left(g\right)=\begin{cases} \phi\left(g\right) & g\in H_{1}\setminus H_{2}\\ \phi^{-1}\left(g\right) & g\in H_{2}\setminus H_{1}\\ g & else \end{cases}$$ but then I'm unable to prove injectivity, for example when $x\in H_{1}\setminus H_{2},y\in H_{1}\cap H_{2}$ we get $\sigma\left(x\right)=\phi\left(x\right),\sigma\left(y\right)=y$ so it is unclear why $\sigma\left(x\right)\neq\sigma\left(y\right)$.

Right now I'm questioning the truthfulness of the claim. But perhaps it's just my constructive proof that's wrong?

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    $\begingroup$ So by the symmetries of $G$, do you just mean the set of bijective maps from $G$ to itself? $\endgroup$ Jan 7, 2017 at 21:17
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    $\begingroup$ I assume it means an automorphism of $G$. At first it seems feasible (and I'd like it to be true, just for aesthetics) but we might run into problems if $H_1$ and $H_2$ "sit differently" in $G$. For example, $S_4$ has a couple Klein four subgroups, but only one is normal in $S_4$; this seems like a source of potential problems. $\endgroup$
    – pjs36
    Jan 7, 2017 at 21:21
  • $\begingroup$ It is true that my original intent was to expand it to a symmetry. However, now that I think of it the answer to this is rather trivial (since the groups are finite, we may use a counting argument). @pjs36 's version is much more interesting! I'll edit my question promptly. $\endgroup$
    – 8l2s
    Jan 7, 2017 at 21:23
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    $\begingroup$ With automorphisms, the answer is "no". There are groups which have isomorphic subgroups where one of them is characteristic (and hence this one can never be send to the other using an automorphism). $\endgroup$ Jan 7, 2017 at 21:26
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    $\begingroup$ I think you can go as small as $D_4$ (dihedral group s.t. $a^4 = b^2 = 1$) to find a counterexample: $Z(D_4) = \langle a^2\rangle \cong \langle b\rangle$ ($\cong \mathbb{Z}_2$), but that isomorphismism clearly can't extend to a group automorphism since the center is characteristic. $\endgroup$
    – sTertooy
    Jan 7, 2017 at 21:35

2 Answers 2


No: e.g., if $H_1$ is a subgroup of the centre of $G$ and $H_2$ is not, then there can be no automorphism of $G$ mapping $H_1$ to $H_2$.

Thanks to Gerry Myerson and SteamyRoot for supplying specific examples.

The conjecture also fails in abelian groups. E.g., take $G = \Bbb{Z}_2 \times \Bbb{Z}_4$ and take $H_1$ and $H_2$ to be the subgroups generated by $x = (0, 2)$ and $y = (1, 0)$ respectively. Then the isomorphism $\phi : H_1 \to H_2$ that maps $x$ to $y$ cannot be extended to $G$ because $x$ is divisible by $2$ but $y$ is not.

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    $\begingroup$ For example, in the group of symmetries of the square, there are five elements of order two. One of them, the rotation halfway 'round, is central, and the group it generates is isomorphic to the group generated by any one of the other elements of order 2 (you just get a cyclic group of order two), but the isomorphism can't be extended to a group automorphism. $\endgroup$ Jan 7, 2017 at 21:35

It's actually kind of hard to get even an automorphism of a subgroup that extends to an automorphism of the entire group (and when that is possible, there are usually many ways to do the extension part). There are some pretty strong conditions that are necessary for what you're asking about (isomorphism of subgroups extending to automorphism of whole group) to even be possible. First, there has to be an automorphism of $G$ which maps the subgroup $H_1$ to $H_2$, which requires that the subgroup lattice 'looks the same' (up to isomorphism types) from the perspective of either subgroup (giving a technical description of what 'looks the same' actually is, is a bit difficult and tedious, but if you look at enough reasonably small subgroup lattices, you should be able to intuit what all has to happen). Moreover, the 'looks the same' condition carries over to the subgroups of $H_1$ and their corresponding subgroups for $H_2$ (which limits which isomorphisms between them will work when it even is possible to do this).


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