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Math Help - Isomorphism Proof

  1. #1
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    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.
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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
    (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?
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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 \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?
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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 \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:

    Quote Originally Posted by Chris L T521 View Post
    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)
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