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Let $H_1$ and $H_2$ be subgroups of some group $G$. Prove that the left $G$-sets $G/H_1$ and $G/H_2$ are isomorphic (as left $G$-sets) iff the subgroups $H_1$ and $H_2$ are conjugate.

If $H_1$ and $H_2$ are conjugate, then they are isomorphic, thus $G/H_1 \cong G/H_2$. I am having problems with the other direction!

Matt Costa
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    Could you please add the definition of "Isomorphic $;G$ - sets" ? – DonAntonio Nov 24 '13 at 20:10
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    Two $G$-sets $X$ and $Y$ are called isomorphic if there is a bijection $\varphi : X\to Y$ with $\varphi(gx) = g\varphi(x)$ for all $g\in G$ and all $x\in X$. – azimut Nov 24 '13 at 21:22
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    be careful because two subgroups can be isomorphic without the corresponding quotients being isomorphic, so in your proof you should use that it isn't just any old isomorphism. –  Nov 24 '13 at 22:28
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    to expand on @Jason's comment, consider for example $G=\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$ with subgroups $H_1=\mathbb{Z}/2\mathbb{Z}\oplus 1$ and $H_2=1\oplus \langle 2+4\mathbb{Z} \rangle$. Then $G/H_1\cong \mathbb{Z}/4\mathbb{Z}$, whereas $G/H_2\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$. – Alexander Gruber Nov 25 '13 at 04:27
  • Have you considered the action of the group $G$ on the set of subgroups via conjugation? –  Nov 24 '13 at 21:37
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    No, however I am having trouble understanding why that would be useful? – Matt Costa Nov 24 '13 at 21:40
  • Ah, yes. I realized this after reading Jason's comment and had the same counterexample actually. Thank you!! – Matt Costa Nov 25 '13 at 14:19

1 Answers1

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If $\phi : G/H_1 \to G/H_2$ is a homomorphism of $G$-sets, and $\phi([1])=[g]$, then it follows more generally that $\phi([x])=[xg]$. But $\phi$ should be well-defined, i.e. $y^{-1} x \in H_1$ implies $(y g)^{-1} x g \in H_2$. This reduces to $g^{-1} H_1 g \subseteq H_2$. Conversely, this relation implies that $\phi([x]):=[xg]$ is a well-defined homomorphism of $G$-sets. Similarly, $\phi$ is injective iff $g H_2 g^{-1} \subseteq H_1$. And $\phi$ is automatically surjective. It follows that $G/H_1 \cong G/H_2$ as $G$-sets iff $H_1$ and $H_2$ are conjugated.

As already pointed out in the comments, it is really important to work in the category of $G$-sets here. But there aren't any alternatives anyway. Of course the category of sets is too weak, and the category of groups doesn't make sense since $H_1,H_2$ aren't assumed to be normal.

Martin Brandenburg
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    Sorry, but could you expand a little more? I am a little confused on your notation [g]. – Matt Costa Nov 25 '13 at 14:42
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    Generally, for an equivalence relation $\sim$ on a given set $X$, the notation $\left[x\right]_\sim$ (or simply $\left[x\right]$ when the relation is obvious) is used to denote the equivalence class of $x$. That is $$\left[x\right]=\left{ y\in X\vert x\sim y\right} $$

    In our case, for $H\leq G$ we denote $\left[g\right]=gH$.

    Notice that in the above answer the coset is of $H_1$ in some usages and of $H_2$ in others. So for example $$\phi\left(\left[1\right]\right)=\left[g\right]$$ should be interpreted as $\phi\left(1H_{1}\right)=gH_{2}$

     – user401346 Jan 08 '17 at 13:23