How to verify that we can define a group operation by putting $a \star b = \phi ^{-1}(\phi (a) \phi (b))$

Question

let $E$ be a set, and $(G, \cdot)$ a group. We suppose that there exists a bijection $\phi : E \rightarrow G$. Show that we can define a law of a group $\star$ on E by: $$\forall a \in E, \forall b \in E, a \star b = \phi ^{-1}(\phi (a) \phi (b)) $$

So basically, I figure I have to show that $(E,\star)$ be a group.

My attempt:

• Associativity: let $a,b,c \in E$, then $$ \begin{align} (a \star b) \star c &= \phi ^{-1}(\phi (a \star b) \phi (c))\\ &= \phi^{-1}(\phi (\phi ^{-1}(\phi (a)\phi(b)) \phi(c))\\ & = \phi^{-1}(\phi(a) \phi(b) \phi(c)) \end{align}$$ and $$\begin{align} a \star (b \star c) &= \phi ^{-1}(\phi (a) \phi (b \star c)) \\ &= \phi ^{-1} (\phi(a)\phi (\phi^{-1}(\phi(a)\phi(b)))) \\ &= \phi^{-1}(\phi(a)\phi(b)\phi(c)). \end{align}$$
• Closure: if $a, b \in E$ then $a \star b \in E$. As $a\star b = \phi ^{-1}(\phi(a) \phi(b))$ and $\phi (a) \in G$, $\phi (b) \in G$ and as G is a group, then $\phi(a) \phi(b) \in G$, thus $\phi ^{-1} (\phi(a) \phi(b)) \in E$
• Identity element: Let $e \in G$ be the identity element of $G$, and let's prove that $\phi ^{-1}(e)$ is the identity element of $E$. We have $\forall a \in G$, $a \star \phi ^{-1}(e)= \phi ^{-1}(\phi (a) \phi (\phi ^{-1}(e)))= \phi ^{-1}(\phi(a) e)=\phi ^{-1}(\phi (a))=a$. Similar proofs goes in showing $\phi^{-1}(e) \star a =a$
• Inverse element: For $a\in G$, then inverse would be the inverse of $\phi (a)$ in $G$, which well denote as $phi(x)$. Proof: $a\star x = \phi^{-1}(\phi (a) \phi(x') )= \phi^{-1}(e)$. Same goes for $x \star a$

Is my proof correct?

Answer 1

Yes, your proof is correct.

As is customary for a proof-verification question, I am required to elaborate on my answer with some feedback on the proposed proof. As such, I will commend you on your decision to split the associativity proof in half, since, otherwise, it would be a sea of symbols to most readers.