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$\Aut(G)$ is the group of isomorphisms from $G$ to $G$ under composition.

For $\Zn{n}$, the automorphism group is $\Zn{n}^{\times}$, the multiplicative group of elements relatively prime to $n$.

For $\Zn{n}^{k}$ the automorphism group is $\GL_{n}(\Zn{n})$, the invertible $n\times n$ matrices with entries in $\Zn{n}$.

The inner automorphisms of $G$ are the conjugations $f_{g}:G\to G$ given by $f_{g}(h)=ghg^{-1}$. The inner automorphisms form a normal subgroup of the automorphism group. The quotient group is called the group of outer automorphisms.

$S_{n}$ has the weird property that all of its automorphisms are inner unless $n=6$.

Conjugacy in $S_{n}$

The conjugacy classes in $S_{n}$ correspond to the cycle decompositions, and there is one class for each partition of $n$ as a sum of positive integers.

The centralizer of a cycle are the permutations which fix the integers appearing in the cycle.

The normalizer of a cycle was computed in the homework, at least in one case.

$A_{5}$ is a simple group

The conjugacy classes in $S_{5}$ that contain even permutations are contained in $A_{5}$ but they might split up into multiple classes.

  • there is one conjugacy class of 3-cycles in $A_{5}$ (there are 20 of these)
  • there are two conjugacy classes of 5-cycles in $A_{5}$ (there are 4!=24 5 cycles, but they split into two groups of 12)
  • all elements of order $2$ in $A_{5}$ are conjugate to $(12)(34)$. There are $15$ of these.

So the conjugacy classes in $A_{5}$ have orders 1, 12, 15, and 20.

If $H$ were a normal subgroup, it would have to have order dividing 60, or 1,2,3,4,5,6,10,12,15,20,30,60. And it would have to be 1 plus a sum of some subset of 12,12,15,20. The only way that works is if it has order 1 or 60.

Proof of the Sylow theorems

See the main page for this section.

Discussion of proof of Sylow’s theorems