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Math Help - Quick quotient group Q

  1. #1
    Senior Member Danneedshelp's Avatar
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    Quick quotient group Q

    I have a simple question about quotient group operations. To this end, if K is a normal subgroup of G, then G/K is group under the binary operation defined by (g_{1}K)*(g_{2}K)=(g_{1}g_{2})K. This is all I have in my notes.

    So, my question is: if the “mother group” G is an additive group, then do we have that (g_{1}K)*(g_{2}K)=(g_{1}+g_{2})K?

    Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    I have a simple question about quotient group operations. To this end, if K is a normal subgroup of G, then G/K is group under the binary operation defined by (g_{1}K)*(g_{2}K)=(g_{1}g_{2})K. This is all I have in my notes.

    So, my question is: if the “mother group” G is an additive group, then do we have that (g_{1}K)*(g_{2}K)=(g_{1}+g_{2})K?

    Thanks
    Let me say it this way. Forget our bad notation for binary operations and let G be a group with binary operation g_1,g_2)\mapsto *(g_1,g_2)" alt="*:G\times G\to Gg_1,g_2)\mapsto *(g_1,g_2)" /> then if N\lhd G we define the binary operation \star:\left(G/N\right)\times\left(G/N\right)\to G/N:\left(g_1N,g_2N\right)\mapsto *(g_1,g_2)N. See if that answeres your question.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    (g_{1})*(Hg_{2}H)=*(g_{1},g_{2})H under the map you defined. So, I would think this means *(g_{1},g_{2})=(g_{1}+g_{2}), because *(g_{1},g_{2}) must be in G, so it follows that * must be the binary operation associated with G.

    Correct?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    (g_{1})*(Hg_{2}H)=*(g_{1},g_{2})H under the map you defined. So, I would think this means *(g_{1},g_{2})=(g_{1}+g_{2}), because *(g_{1},g_{2}) must be in G, so it follows that * must be the binary operation associated with G.

    Correct?
    Hmm...close enough. My point was that the notation for the binary operation is irrelevant. So, in words the binary operation on G/N is described "The (insert binary operation here) of two cosets is the coset whose coefficient (you get what I mean) is the (insert binary operation here) of the coefficients of the two original cosets"
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