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Thread: Isomorphism Proof

  1. #1
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    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.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by demode View Post
    $\displaystyle (G, \circ)$ 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?
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  3. #3
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    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?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by demode View Post
    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:

    Quote Originally Posted by Chris L T521 View Post
    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)
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