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**demode** $\displaystyle (G, \circ)$ is a group. The operation $\displaystyle \star$ on $\displaystyle G$ is defined by $\displaystyle a \star b = b \circ a$ for all $\displaystyle a,b \in G$. Let $\displaystyle \phi$ be the map from G to itself such that $\displaystyle \phi(g)=g^{-1}$ where $\displaystyle g^{-1}$ is the inverse of $\displaystyle g$ in group $\displaystyle (G, \circ)$. Show that $\displaystyle \phi$ is a group isomorphism from $\displaystyle (G, \star)$ to $\displaystyle (G, \circ)$.

__Attempt:__

I think I have to show that $\displaystyle \phi$ is one to one, onto and it preserves order:

For "one to one":

$\displaystyle g_1^{-1} = g_2^{-1}$

$\displaystyle \iff (b_1 \circ a_1) = (b_2 \circ a_2)$

$\displaystyle \iff (a_1 \star b_1) = (a_2 \star b_2)$

$\displaystyle \iff g_1 = g_2$

For "onto":

I must find some $\displaystyle y \in G$ such that $\displaystyle \phi(g)=y=g^{-1}$.

$\displaystyle \phi (a \star b) = b \circ a$

So $\displaystyle y= b \circ a$

And to show that it's operation preserving I did the following:

$\displaystyle \phi (g_1,g_2) = \phi ((a_1 \star b_1), (a_2 \star b_2))$

$\displaystyle =g_1^{-1}g_2^{-1} = (b_1 \circ a_1)(b_2 \circ a_2) = \phi(g_1)\phi(g_2)$

I appreciate it if anyone could correct any of my mistakes.