1. ## Permutation groups

1a) Prove that for any two 3-cycles $\sigma_1, \sigma_2 \in S_n (n>4)$ there exists a $\tau \in A_n$such that $\tau\sigma_1\tau^{-1}=\sigma_2$
1b) Let $Z(S_n)$ be the center of $S_n$. Prove that $Z(S_n)=Z(A_n)=\{e\}$ for all $n>3$ and $Z(S_3)=\{e\}$
1c) Prove that the only nontrivial normal subgroup of $S_n (n>4)$ is $A_n$
1d) Give a nontrivial normal subgroup $N$ of $S_4$ such that $N \ne A_4$
can anyone help? thanks in advance

2. Originally Posted by tszhin8831
1a) Prove that for any two 3-cycles $\sigma_1, \sigma_2 \in S_n (n>4)$ there exists a $\tau \in A_n$such that $\tau\sigma_1\tau^{-1}=\sigma_2$
If $\sigma_1 = (a,b,c)$ then $\tau \sigma \tau^{-1} = (\tau (a), \tau(b) , \tau (c))$.
Therefore, if $\sigma_2 = (d,e,f)$ then let $\tau$ be defined as $a\mapsto d, b\mapsto e, c\mapsto f$ and fixes everything else.
The problem is that $\tau = (a,d)(b,e)(c,f)\not \in A_n$.
In order to fix that choose $a'\not \in \{a,b,c\}$ and $b'\not \in \{a',d,e,f\}$ and redefine $\tau$ to send $a'\mapsto b'$.
This is possible since $n>4$. And we see that $\tau = (a,d)(b,e)(c,f)(a',b')\in A_n$.

3. Originally Posted by tszhin8831
1b) Let $Z(S_n)$ be the center of $S_n$. Prove that $Z(S_n)=Z(A_n)=\{e\}$ for all $n>3$ and $Z(S_3)=\{e\}$
Let $\sigma\in Z(S_n)$ if $\sigma_1$ is conjugate to $\sigma$ then it means $\tau \sigma \tau^{-1} = \sigma_1 \implies \sigma = \sigma_1$.
Therefore, elements in the center are only conjugate to themselves.
We will show that any non-identity element in $S_n$ ( $n>2$) are conjugate to some other element.
Let $\theta \in S_n$ be a cycle, $\theta = (a_1,...,a_k)$, $k\geq 2$.
Then $\tau \theta \tau^{-1} = (\tau(a_1), ... , \tau(a_k))$.
Therefore if we pick $\tau = (a_2,...,a_k,a_1)$ then $\tau \theta \tau^{-1} \not = \theta$ if $k>2$.
If $k=2$ then $(a_1,a_2)$ and $(a_2,a_1)$ are the same thing.
But in this special case we can just let $\tau = (a_1,a_3)$ where $a_3 \not \in \{a_1,a_2\}$ and so $\tau \theta \tau^{-1} \not = \theta$.
We have shown that no cycle can be in $Z(S_n)$ for $n>2$.

Let $\sigma \in S_n$ be a non-identity permutation.
Then we can write $\sigma = \theta_1 \theta_2 ... \theta_j$ into a product of disjoint cycles.
Therefore, $\tau \sigma \tau^{-1} = ( \tau \theta_1 \tau^{-1})(\tau \theta_2 \tau^{-1}) ... (\tau \theta_j \tau^{-1})$.
For the moment assume there is one of $\theta_1,...,\theta_j$ that is not a transposition, say, $\theta_1$ by relabing.
Then it means, $\theta_1 = (x_1,...,x_r)$ where $r>2$.
Let $\tau = (x_2,...,x_r,x_1)$ and we see that $\tau \theta_i \tau^{-1} = \theta_i$ for $i\not = 1$.
Therefore, $\tau \sigma \tau^{-1} = \tau \theta_1 \tau^{-1} \theta_2 ... \theta_j$ while $\sigma = \theta_1 ... \theta_j$ so $\tau \sigma \tau^{-1} \not = \sigma$ since $\tau \theta_1 \tau^{-1} \not = \theta_1$.
We have constructed an element conjugate and not equal to $\sigma$ thus $\sigma \not \in Z(S_n)$.
But we left the case when each $\theta_i$ is a transposition as an exceptional case.
Try to see what you can come up with to complete the proof for $Z(S_n) = \{ e \}$ for $n>2$.

As for $Z(A_n)$ the proof should be similar but needs to be done with more care.

4. Originally Posted by tszhin8831
1c) Prove that the only nontrivial normal subgroup of $S_n (n>4)$ is $A_n$
If you happen to know that $A_n$ is simple for $n>4$ then that is really helpful here. Accepting this fact we see that $\{ e \} \subset A_n \subset S_n$ is a composition series. By Jordan-Holder theorem this is the unique composition series. Consequently, it must mean that $A_n$ is the only proper non-trivial normal subgroup of $S_n$.
1d) Give a nontrivial normal subgroup $N$ of $S_4$ such that $N \ne A_4$
can anyone help? thanks in advance
Choose $N = \{ \text{id}, (12)(34),(13)(24),(14)(23)\}$.

Here is another related problem can you show that $N$ and $A_4$ are the only proper non-trivial normal subgroups of $S_4$?

5. thanks a lot