Normal subgroup of group under symmetric group

Question

Given $\rho=(3,7)$ and $\sigma=(1,7,3)(2,10,4,8)$ where $\rho,\sigma\in S_{10}$

Let $\langle\sigma,\rho\rangle$ be the subgroup of $S_{10}$ generated by $\sigma$ and $\rho$

Is the cyclic subgroup $\langle\sigma\rangle$ generated by $\sigma$ a normal subgroup of $\langle\sigma,\rho\rangle$ ?

$\mathbf{Generally:}$ A normal subgroup $N$ of a group $G$ must satisfy that for each element $n \in N$ and each $g \in G$, the element $gng^{-1}$ is still in $N$

I'm thinking that I have to try to find a case where this is not true, but fear that this can be a lot of calculation.

Can I avoid a lot of calculations in this case? Does it help me knowing that $\rho\sigma\rho^{-1} = (1,3,7)(2,10,4,8)$ ?

$\mathbf{Edit}$:

I noticed that $\sigma^5 = (1,3,7)(2,10,4,8)$

Does this mean it is a normal subgroup?

Answer 1

Hint:

If a group is given by its generators $\;G=\langle\,a,b,c,...\rangle\;$ , then the cyclic subgroup generated by anyone of those generators in normal in $\;G\;$ iff conjugating it by the other generators is a power of that first generator, meaning:

$$\langle a\rangle\lhd G\iff a^x:=x^{-1}ax\in\langle a\rangle\;,\;\;\forall\;x\in\{b,c,...\}$$

Answer 2

Another way of looking at it: as you remarked $\rho\sigma\rho^{-1}=\sigma^5$, and observe that the order of $\sigma \text{ is } 12$, and that of $\rho$ is $2$. This shows that $\langle\sigma,\rho\rangle= \{\sigma^i\rho^j \mid i=0,\dots,11 \text{ and } j=0,1\}$ (in fact a subgroup of order $24$ of $S_{10}$). Clearly index$[\langle\sigma,\rho\rangle:\langle\sigma\rangle]=2$, and a basic fact gives us subgroups of index $2$ are normal.