1. ## Isomorphism Proof

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

Attempt:

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

For "one to one":

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

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

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

$\iff g_1 = g_2$

For "onto":

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

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

So $y= b \circ a$

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

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

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

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

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

Attempt:

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

For "one to one":

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

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

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

$\iff g_1 = g_2$

For "onto":

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

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

So $y= b \circ a$

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

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

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

I appreciate it if anyone could correct any of my mistakes.
I would approach it this way:

For injectivity, show that $\phi(g_1)=\phi(g_2)\iff g_1=g_2$.

So $\phi(g_1)=\phi(g_2)\iff g_1^{-1}=g_2^{-1}$.

Since $g_1,g_2\in (G,\star)$, it follows that $g_1^{-1}=g_2^{-1}\iff (g_1^{-1})^{-1}=(g_2^{-1})^{-1}\iff g_1=g_2$.

For sujectivity, we seek a $x \in G: \phi(x)=y=g^{-1}$, in this case, we can say $x=g$.

Now, let $a,b\in (G,\star)$. Then

\begin{aligned}\phi(a\star b) &= (a\star b)^{-1}\\ &= b^{-1}\star a^{-1}\\ &=a^{-1}\circ b^{-1}\\ &=\phi(a)\circ \phi(b)\end{aligned}

Thus $\phi: (G,\star) \rightarrow (G,\circ)$ is an isomorphism (more importantly, this is an automorphism)

Does this make sense?

3. Thank you. But I think to show that it's isomorphism, we also need to show that $\phi$ is operation-preserving; that is to show that $\phi(ab)=\phi(a)\phi(b) \forall a,b \in G$. In my first post I tried to show this condition. Was it correct?

4. Originally Posted by demode
Thank you. But I think to show that it's isomorphism, we also need to show that $\phi$ is operation-preserving; that is to show that $\phi(ab)=\phi(a)\phi(b) \forall a,b \in G$. In my first post I tried to show this condition. Was it correct?
I couldn't quite follow what you were doing, so to me it doesn't seem correct. If you didn't notice, I did the operation preserving part in my reply to your question:

Originally Posted by Chris L T521
Now, let $a,b\in (G,\star)$. Then

\begin{aligned}\phi(a\star b) &= (a\star b)^{-1}\\ &= b^{-1}\star a^{-1}\\ &=a^{-1}\circ b^{-1}\\ &=\phi(a)\circ \phi(b)\end{aligned}

Thus $\phi: (G,\star) \rightarrow (G,\circ)$ is an isomorphism (more importantly, this is an automorphism)