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This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 $\rightarrow$ 2, 2 $\rightarrow$ 1 and 3 $\rightarrow$ 4, 4 $\rightarrow$ 3) taken over the three possible such combinations of four elements.

Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to $S_3$. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help.

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  • $\begingroup$ I don't understand the question. What do you mean by "what does it mean to have a permutation of these three permutations"? $\endgroup$ – Qiaochu Yuan Jan 28 '12 at 21:13
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    $\begingroup$ Call the three non-identity elements $x, y, z$. Then a permutation is just a permutation of the set containing $x,y,z$. Perhaps you are confused because $x,y,z$ can themselves be written as permutations, but one ignores the detailed structure of $x,y,z$ and just treats them as items that can be permuted. $\endgroup$ – Gerry Myerson Jan 28 '12 at 22:07
  • $\begingroup$ @GerryMyersonThanks. That's actually my question. Using the description of {1,2,3,4} and the permutations as I mentioned, what explicitly happens to "1" for example if you call 1 $\rightarrow$ 2 etc. $x$ and call 1 $\rightarrow$ 3, etc. $y$ and you send $x$ to $y$? $\endgroup$ – user12802 Jan 28 '12 at 22:16
  • $\begingroup$ @QiaochuYuanI mentioned in the comment to Gerry what I am trying to articulate. If you have these permutations representing the Klein 4-group, What happens to them as you permute the permutations amongst themselves. $\endgroup$ – user12802 Jan 28 '12 at 22:20
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Every group is a set (together with a binary operation on the set). Every automorphism of a group $G$ is a bijective function from the underlying set of $G$ to itself (which in addition respects the operation of $G$). So every automorphism of a group $G$ can be viewed as a subgroup of the group of all permutations on the underlying set of $G$; in fact, since an automorphism must send the identity of $G$ to itself, you can even view every automorphism of $G$ as a permutation of the set $G-\{e\}$.

Here you are getting a bit confused because you are viewing your group $G$ as a subgroup of $S_4$ (nothing wrong with that), and then you are trying to understand $\mathrm{Aut}(G)$ (which can be viewed as a subgroup of $S_{G-{e}} \cong S_3$) as acting on the set $\{1,2,3,4\}$ as well. While some automorphisms can be defined in terms of an action on $\{1,2,3,4\}$, not every automorphism can.

If you view $G$ as the set $\{\mathrm{id}, (1\;2), (3\;4), (1\;2)(3\;4)\}$, then letting $$\begin{align*} x &= (1\;2),\\ y &= (3\;4),\\ z &= (1\;2)(3\;4)\\ e &= \mathrm{id} \end{align*}$$ then the automorphisms of $G$ will always map $e$ to itself, and so you can view the automorphisms as being elements of $S_{\{x,y,z\}}$, the permutation group of $\{x,y,z\}$. The elements of the automorphism group are then (written in 2-line format): $$\begin{align*} \mathrm{id}_{\{x,y,z\}} &= \left(\begin{array}{ccc} x & y & z\\ x & y & z \end{array}\right)\\ f_1 &= \left(\begin{array}{ccc} x & y & z\\ y & z & x \end{array}\right)\\ f_2 &=\left(\begin{array}{ccc} x & y & z\\ z & x & y \end{array}\right)\\ f_3 &= \left(\begin{array}{ccc} x & y & z\\ y & x & z \end{array}\right)\\ f_4 &= \left(\begin{array}{ccc} x & y & z\\ z & y & x \end{array}\right)\\ f_5 &= \left(\begin{array}{ccc} x & y & z\\ x & z & y \end{array}\right) \end{align*}$$ and you can verify that each of them is an automorphism; since every automorphism corresponds to a unique element of $S_{\{x,y,z\}}$, and every element of this group is an automorphism, then this is the automorphism group.

However, you are trying to view your group $G$ as a subgroup of $S_4$. Can the automorphisms of $S_4$ be "induced" by some permutation of $\{1,2,3,4\}$? Equivalently:

If we view the Klein $4$-group $G$ as the subgroup of $S_4$ generated by $(1\;2)$ and $(3\;4)$, is every automorphism of $G$ induced by conjugation by an element of $S_4$?

Well, it doesn't. The reason it doesn't is that $x$, $y$, and $z$ don't all have the same cycle structure: conjugation by elements of $S_4$ (or more precisely, by elements of the normalizer of $G$ in $S_4$) will necessarily map $z$ to itself, because $z$ is the only element with its cycle structure (product of two disjoint transpositions) in $G$. So the only automorphisms that can be viewed as coming from "acting on ${1,2,3,4}$" are the identity and $f_3$

Added: However: it is possible to view the Klein $4$-group as a different subgroup of $S_4$: identify $x\mapsto (1\;2)(3\;4)$; $y\mapsto (1\;3)(2\;4)$, $z\mapsto (1\;4)(2\;3)$. It is not hard to verify that this is a subgroup of order $4$, and since it has three elements of order $2$, it is isomorphic to the Klein $4$-group. Now you can indeed realize every automorphism of $G$ via conjugation by an element of $S_4$ (in several different ways). Here's one way: the automorphism $\mathrm{id}$ can be realized as conjugation by the identity. The automorphism $f_1$ is given by conjugation by $(2\;3\;4)$; $f_2$ is given by conjugation by $(2\;4\;3)$; the automorphism $f_3$ is given by conjugation by $(2\;3)$; $f_4$ is given by conjugation by $(2\;4)$; and $f_5$ is given by conjugation by $(3\;4)$. You can now easily see that the automorphism group is indeed isomorphic to $S_3$, since it corresponds precisely to the subgroup of $S_4$ that fixes $1$ (you can take any of the other four one-point stabilizers as well).

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    $\begingroup$ @ArturoMagidinThanks. And congratulations for passing the 100k milestone. In addition to giving great answers, you always seem to have the insight/patience to write at the level of the questioner. $\endgroup$ – user12802 Jan 28 '12 at 22:56
  • $\begingroup$ @Andrew: I just added something you might find closer to what you were thinking about. And thanks. $\endgroup$ – Arturo Magidin Jan 28 '12 at 23:04
  • $\begingroup$ @ArturoMagidinThanks again. That was what I had a problem with - as I was starting with the x,y,z you just added but then couldn't see what to do with them to get the automorphisms. $\endgroup$ – user12802 Jan 28 '12 at 23:59

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