1. ## Isomorphism Proof

$\displaystyle (G, \circ)$ 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.

2. Originally Posted by 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.
I would approach it this way:

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

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

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

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

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

\displaystyle \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 $\displaystyle \phi: (G,\star) \rightarrow (G,\circ)$ 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 $\displaystyle \phi$ is operation-preserving; that is to show that $\displaystyle \phi(ab)=\phi(a)\phi(b) \forall a,b \in G$. 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 $\displaystyle \phi$ is operation-preserving; that is to show that $\displaystyle \phi(ab)=\phi(a)\phi(b) \forall a,b \in G$. 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 $\displaystyle a,b\in (G,\star)$. Then

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