Showing something is a subgroup

If $\sigma_1$ , $\sigma_2$ are permutations which fix an element $a \in X$ , $\sigma_i \in S_X$ , then you need to prove two things,

i. $\sigma_1^{-1}$ also fixes $a$ .

ii. $\sigma_1 \sigma_2$ also fixes $a$ .

Can you work out why these two things are sufficient?

Now, to prove i. you should note that $\sigma_1: a \mapsto a$ . If $\sigma^{-1}: a \mapsto b$ then where must $b$ be sent to in $(\sigma_1^{-1})^{-1} = \sigma_1$ ?

To prove ii. simply plug in $a$ into $\sigma_1 \sigma_2$ .