Here is an exercise from [Birkhoff and MacLane, A Survey of Modern Algebra]:

Let $\phi: G \rightarrow G'$ be an isomorphism between two groups of permutations. Let $S$ consist of those permutations of $G$ leaving one letter fixed. Does its image $S'$ form a subgroup of $G'$? Must the set $S'$ leave a letter fixed? Illustrate.

I can show that $S'$ forms a subgroup of $G'$. I was wondering about whether $S'$ must leave a letter fixed. I think the answer is a `no' in general, and here is one solution, though I'd be glad to know of other approaches.

We can let $G$ be an embedding of $S_3$ into $S_4$, and $G'$ an embedding of $S_3$ into $S_6$, as follows. $G:= \langle (a,b,c)(d), (a,b)(d) \rangle$. To define $G'$, let $r=(1,2,3,4,5,6),s=(2,6)(3,5)$ (these are the usual rotation and reflection in the symmetries of the regular hexagon). Take $G' := \langle rs,r^2 \rangle = \langle (1,2)(3,6)(4,5),(1,3,5)(2,4,5)\rangle$. Then $G'$, which is a subgroup of the dihedral group $D_{12}$, is isomorphic to $S_3$ (it is a transitive embedding of $S_3$ into $S_6$). Thus, $G \cong G'$, and an explicit isomorphism is given by $\phi: (a,b) \mapsto rs, (a,b,c) \mapsto r^2$.

If we let $S$ be the stabilizer in $G$ of $d$, then $S$ equals all of $G$ since $G$ itself fixes $d$. And $S'$ equals all of $G'$ and hence has no fixed points. This gives an example where $S$ fixes a point but $S'$ doesn't.

If we let $S$ be the stabilizer in $G$ of $c$, then $S = \langle (a,b) \rangle$, and $S' = \langle rs \rangle = \langle (1,2)(3,6)(4,5) \rangle$. Here too, $S$ fixes a point but $S'$ doesn't.

Of course, if the two permutation groups $G,G'$ are identical (which implies that their degrees are equal), then the image of a stabilizer would be another stabilizer.

  • $\begingroup$ Seems fine. If $\phi$ is an isomorphism of permutation groups (so not only are $G$ and $G'$ of the same degree, but they act in the same way, just on different things), then of course $S'$ fixes the point corresponding the point fixed by $S$. If $\phi$ is only an isomorphism of groups (but $G$ and $G'$ happen to be permutation groups), then your examples are clear and correct. $\endgroup$ – Jack Schmidt Jan 2 '14 at 1:23

How about $S = \{1,2,3,4\}$, $G_1 = \{\mathrm{id}, (24)\}$, $G_2 = \{\mathrm{id}, (13)(24)\}$? $G_1$ and $G_2$ are isomorphic, and $G_1$ fixes $1$ and $3$ but $G_2$ doesn't fix anything.

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