Prove $(S,\cdot)$ is a group.

Question

Let $G$ be a group and $H \leq \text{Aut}(G)$. On Cartesian product $S=G \times H$ define multiplication: $$ \text{For} \,\, a,b \in G \,\, \text{and} \,\, \alpha,\beta\in \text{Aut}(G) \,\, \text{let} \,\, (a,\alpha)\cdot(b,\beta)=(a\alpha^{-1}(b),\alpha \beta).$$

Prove $(S,\cdot)$ is a group.

In my attempt I wasn't sure if I was taking the proper approach so I left out the identity.

$\textbf{Proof:}$ Since $G$ is a group and $H \leq \text{Aut}(G)$ is a group, then $\forall a,b \in G, \,\,\, a \cdot b \in G$ by closure. So, $\alpha, \beta \in H \implies \alpha, \beta \in \text{Aut}(G)$. Thus, $\alpha \cdot \beta \in H,\text{Aut}(G)$ by closure. So, $(a,\alpha) \cdot (b,\beta)=(a\alpha^{-1}(b),\alpha \beta) \in S \,\, \forall (a, \alpha),(b,\beta) \in S$.
$\therefore S$ is closed under multiplication.
For associativity, we want to show $((a,\alpha) \cdot (b,\beta)) \cdot (d, \delta)=(a,\alpha) \cdot ((b,\beta)\cdot(d,\delta)) \,\,\, \forall (a,\alpha),(b,\beta),(d,\delta) \in S$.
$((a,\alpha) \cdot (b,\beta)) \cdot (d,\delta) = (a\alpha^{-1}(b),\alpha \beta) \cdot (d,\delta)=(a\alpha^{-1}(b)(\alpha \beta)^{-1}(d),\alpha \beta \delta)$
$(a,\alpha)\cdot ((b,\beta)\cdot (d,\delta))=(a\alpha^{-1}(b\beta^{-1}(d),\alpha \beta \delta)$ At this point I am not sure if I am on the right track and need to show these last two steps are equivalent.

Need to show there exists inverse and identity for $S$ which I said is true since it exists for groups $G$ and $H$.

Answer 1

• Well-definedness (closure): Given $a,b \in G$ and $\alpha,\beta \in H$, obviously $a\alpha^{-1}(b) \in G$ and $\alpha \circ \beta \in H$ since $H$ is a subgroup of $\operatorname{Aut}(G)$. Thus $(a,\alpha) \cdot (b,\beta) = (a\alpha^{-1}(b),\alpha \circ \beta) \in G \times H$.
• Associativity: Given $a,b,c \in G$ and $\alpha,\beta,\gamma \in H$, $$(a\alpha^{-1}(b),\alpha \circ \beta) \cdot (c,\gamma) = (a\alpha^{-1}(b)(\alpha \circ \beta)^{-1}(c),(\alpha \circ \beta) \circ \gamma)$$ and, on the other hand $$(a,\alpha) \cdot (b\beta^{-1}(c),\beta \circ \gamma) = (a\alpha^{-1}(b\beta^{-1}(c)),\alpha \circ (\beta \circ \gamma)) = (a\alpha^{-1}(b)\alpha^{-1}(\beta^{-1}(c)),\alpha \circ (\beta \circ \gamma)),$$ so, they are the same.
• Identity: If $1_G$ is the identity of $G$, just consider $(1_G,\operatorname{id}_G) \in G \times H$, and check that $$(a,\alpha) \cdot (1_G,\operatorname{id}_G) = (a,\alpha) = (1_G,\operatorname{id}_G) \cdot (a,\alpha).$$
• Inverses: Given $a \in G$ and $\alpha \in H$, check that $$(a,\alpha) \cdot (\alpha(a^{-1}),\alpha^{-1}) = (\alpha(a^{-1}),\alpha^{-1}) \cdot (a,\alpha) = (1_G,\operatorname{id}_G).$$