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I've just been learning about Cayley's theorem and a couple of things occurred to me:

We know that every finite group of order $n$ is isomorphic to some subgroup of $S_n$. But perhaps there are multiple subgroups of $S_n$ for which there exists such an isomorphism. So for instance, the Klein four-group is isomorphic to

$$ \{(), (12)(34), (13)(24), (14)(23) \} $$

but also to

$$\{ (), (12), (34), (12)(34)\} \quad \mathrm{and} \quad \{ (), (14), (23), (14)(23)\} \quad \mathrm{and} \quad \{ (), (13), (24), (13)(24)\} \,.$$

So my question is: for a given subgroup of $S_n$, how many other subgroups of $S_n$ are there which are isomorphic to it?

The other thing that occurred to me is that there are often multiple ways to construct an isomorphism, given a subgroup of $S_n$. Taking the cyclic group of order 3, for instance, we could construct a homomorphism $\phi : C_3 = \{ e, c, c^2\}\to S_3$ like so

$$ \phi(e) = () $$ $$ \phi(c) = (123) $$ $$ \phi(c^2) = (132) $$

But it would be just as legitimate to switch the right hand sides of the second and third equations --- we would still have a homomorphism. Which seems to suggest that $c$ and $c^2$ are in some sense equivalent. I thought at first that this might be to do with conjugation (I knew, for instance, that for the dihedral group the reflections in the planes of symmetry are conjugate, and in some sense more similar to one another than to the $n$-fold rotations). Of course, $c$ and $c^2$ are not conjugate (within $C_3$, that is), and so this leads to me to ask the question:

In what sense are $c$ and $c^2$ equivalent; given a group $G$, what can we say about the non-trivial isomorphisms $\phi_i : G \to G$? For a given element $g$, the set of all elements $\phi_i(g)$ must form an equivalence class of some sort --- does this have a name?

I appreciate this is a very general question, so if at best you could point me to some references and further reading, that would be appreciated.

gj255
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  • Actually, your $\phi(c)$ and $\phi(c^2)$ are conjugate in $S_3$. Howver your first representation of $V_4$ is not conjugate to the other three (which are mutually conjugate), because the cycle structures of the elements don't match up. – hmakholm left over Monica Oct 01 '14 at 16:04
  • This leads me to even more questions! For instance, it appears that for a set of subgroups of $S_n$ which are mutually isomorphic, we can partition them into conjugacy classes. How does this partitioning work? Can we tell, just by looking at the group $V_4$, that it will be isomorphic to 4 subgroups of $S_n$, having a 3-1 split in terms of being conjugate? Moreover, what is the relationship between two group elements which are not conjugate within a given group, but are conjugate when considered as part of a bigger group? Can we always find a big enough group that any two will be conjugate? – gj255 Oct 01 '14 at 16:12
  • x @gj255: For the last question: not always. At least your luck is out if the two elements have different orders. – hmakholm left over Monica Oct 01 '14 at 16:17
  • It just occurred to me: is this 3-1 split of the subgroups of $S_4$ which are isomorphic to $V_4$ related to the fact that within $V_4$, $a$, $b$ and $ab$ are in some sense equivalent, whilst $e$ is very much distinct? – gj255 Oct 01 '14 at 16:24
  • You should look at this question, particularly the accepted answer. – rogerl Oct 01 '14 at 16:32
  • On the other hand, if $G$ is finite and you have $a,b\in G$ with the same order and $\phi:G\to S_{|G|}$ is the regular embedding (given by the group's action on itself by multiplication), then $\phi(a)$ and $\phi(b)$ will be conjugate in $S_{|G|}$, since they have the same cycle structure (namely $|G|/|a|$ disjoint $|a|$-cycles). There's a similar result for infinite groups. – hmakholm left over Monica Oct 01 '14 at 16:42

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