I am trying to understand the proof that the only normal subgroups of $S_5$ are $\{1\}$, $A_5$, and $S_5$. The proof starts as follows:
Suppose $H \trianglelefteq S_5$. Restrict $\operatorname{sgn}:S_5 \mapsto\{\pm 1\}$ to $H$. If $H \leq A_5$ then $H \trianglelefteq A_5$ (why??), so $H = \{1\}$ or $A_5$ because $A_5$ is simple. Thus, assume that $H$ contains both even and odd permutations, so that $H \cap A_5 \trianglelefteq A_5$. Thus, $H\cap A_5 = \{1\}$ or $A_5$. In the second case $H=S_5$. In the first case, $H \cong \{\pm 1\}$ (Why? I don't understand the following part) because $\operatorname{sgn}:H \mapsto\{\pm 1\}$ is onto and $A_5 \cap H = \operatorname{Ker}(\operatorname{sgn})\cap H = \{1\}$. Thus, ... (I understand the rest of the proof)
I would appreciate help understand the parts in bold. Thanks!
For the first part, if $H \trianglelefteq S_5$ and we have $H \leq A_5 \leq S_5$, then $H \trianglelefteq A_5$. This is because if $sHs^{-1} \in H$ for each $s \in S_5$, then because $A_5$ is a subset of $S_5$, $aHa^{-1} \in H$ for each $a \in A_5$.
In the second part, I'm not sure what it means for $H$ to be isomorphic to $\{\pm 1\}$ since $\{\pm 1\}$ are just the signs of the permutations and not group elements. Maybe the author is looking at the map $\operatorname{sgn} \colon H \to \{\pm1\}$ and saying that $H \cong \operatorname{Im}\operatorname{sgn}/\operatorname{Ker}\operatorname{sgn}$, and $\operatorname{Ker}\operatorname{sgn}$ has only a single element?