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Let $G$ be a group. For $c \in G$ a fixed element let us define a new operation in $G$ defined by $a*b=a\cdot c \cdot b$

Prove that the funtion $f: (G, \cdot)\longrightarrow (G, *) $ given by $f(x)=c^{-1} \cdot x$ is an isomorphism

my attempt:

$f(x)=c^{-1}x $ $f(y)=c^{-1}y $

$f(x)= f(y): c^{-1}x = c^{-1}y \Longrightarrow{x=y} $

How do I test for surjectivity and morphism?

Thanks

Mathcristh80
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    what did you try for those two? – Quimey Apr 21 '21 at 13:51
  • Not strictly speaking a duplicate of the linked question. That is asking for a proof that $(G,*)$ is indeed a group (with $c^{-1}$ being the identity). This question seems to assume that fact already, and is now asking for a proof that a particular map out of the original group $G$ is an isomorphism. – John Gowers Apr 21 '21 at 14:26

1 Answers1

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As now pointed out, there is the issue of whether the new operation makes $G$ into a group. Suppose it does.

Given $y\in G $, we have $f (cy)=c^{-1}cy=y$, proving surjectivity.

Now for morphism, you have to use the new rule. $f (xy) =c^{-1}xy$. But $f (x)*f (y)=(c^{-1}x)*(c^{-1}y)=c^{-1}xcc^{-1}y=c^{-1}xy=f (xy) $.