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

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?

• Your proof is correct! Commented Sep 26, 2017 at 18:00