Product of permutations and subgroup generated by permutation(s) I'm get confused while working with permutations, so I have some questions. 
$\sigma$ = (1,7,3)(2,10,4,8)
$\rho$ = (3,7)
$\tau$ = (1,7)
First I am told to compute $\tau$$\sigma$$\tau^{-1}$ 
I dont if i went about it the right way, anyway my final result is (1,3,7)(2,10,4,8) (I computed  $\tau$$\sigma$ first - right to left- and then ($\tau$$\sigma$ )$\tau^{-1}$.) Is my result correct? 
<$\sigma$,$\rho$> is a subgroup of $S_{10}$. I have to check if $\sigma$ is a normal subgroup of <$\sigma$,$\rho$>. My idea was to use the index=2 check by finding the order of <$\sigma$,$\rho$> . Will that work? 
Also, what does <$\sigma$,$\rho$> look like? if it were <$\sigma$$\rho$> I understand the elements, but i am unsure whether the comma means something that i am not grasping.  
 A: In theory you can do the multiplication in any order you like, because groups are associative. However, in practice you can actually do all the multiplication simultaneously! The permutation you wrote down is the correct answer, but this is hopefully a quicker method:
You said $\sigma = (1 \: 7 \: 3)(2 \: 10 \: 4 \: 8)$ and $\tau = (1 \: 7)$, and we want to calculate $\tau \sigma \tau^{-1}$. We know this is some permutation on $\{1, \ldots, 10\}$, so can just figure out where it send various elements.
Let's start with 1. $\tau^{-1}$ sends 1 to 7. $\sigma$ sends 7 to 3. $\tau$ doesn't affect 3, so we know $\tau \sigma \tau^{-1}$ sends 1 to 3. If we continue in this way, we find $\tau \sigma \tau^{-1}$ then sends 3 to 7 and 7 to 1, so have $\tau \sigma \tau^{-1} = (1\: 3\: 7)  (\text{some other stuff})$.
(In fact, there's an even quicker way: when we conjugate $\sigma$ by $\tau$ that has the same effect as relabelling the names of the elements that $\sigma$ permutes, by the permutation given by $\tau$. That is, we swap over the 1 and the 7 in the definition of $\sigma$!)

In answer to your next question, if the index of $H$ in $G$ is 2 then $H$ is normal, but not all normal subgroups of $G$ have index 2 I'm afraid.
In order to answer the question about how to check normality, I'll actually clear up your last question first: $\langle \sigma, \rho \rangle$ is the group generated by $\sigma$ and $\rho$. It's the smallest group that contains both $\sigma$ and $\rho$, and consists of all the possible things you can get from multiplying together $\sigma$, $\sigma^{-1}$, $\rho$ and $\rho^{-1}$.
This means that to check if $\langle \sigma \rangle$ is normal in $\langle \sigma, \rho \rangle$, it's enough to check whether when you conjugate $\sigma$ by $\sigma$, $\sigma^{-1}$, $\rho$ and $\rho^{-1}$ you end up in $\langle \sigma \rangle$ in each case.
However, clearly this will be the case for $\sigma$ (because $\langle \sigma \rangle$ is a subgroup), so you don't need to check $\sigma$ or $\sigma^{-1}$. Also, if $\rho \sigma \rho^{-1} \in \langle \sigma \rangle$ then $\rho^{-1} \sigma \rho \in \langle \sigma \rangle$, so you in fact only need to check $\rho \sigma \rho^{-1} \in \langle \sigma \rangle$.
