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Thread: Direct Product

  1. #1
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    Direct Product

    Let G = (G, . , e), H = (H , * , E) be groups ... (e is the identity)
    the direct product is defined by:

    G x H = (GXH, o , (e,E)) where,
    (g1,h1) o (g2,h2) = (g1 . g2, h1*h2)

    Question: Show formally that G x H is a group.
    when it says, "show formally that G x H is a group".. does it mean show that G x H satisfies the 4 conditions for being a group.. i.e.

    associativity,
    closure,
    existance of identity element and
    existance of inverse element?

    any help is appreciated..
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  2. #2
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    Quote Originally Posted by smoothman View Post
    Let G = (G, . , e), H = (H , * , E) be groups ... (e is the identity)
    the direct product is defined by:

    G x H = (GXH, o , (e,E)) where,
    (g1,h1) o (g2,h2) = (g1 . g2, h1*h2)

    Question: Show formally that G x H is a group.
    when it says, "show formally that G x H is a group".. does it mean show that G x H satisfies the 4 conditions for being a group.. i.e.

    associativity,
    closure,
    existance of identity element and
    existance of inverse element?
    First, closure is not a requirement. Because we define a binary operation $\displaystyle *$ such that $\displaystyle *:G\times G\mapsto G$, so of course it is closed. However, a subgroup needs to satisfy the closure property.

    If you define $\displaystyle *G\times H)\times (G\times H)\mapsto G\times H$ as $\displaystyle (g_1,h_1)*(g_2,h_2) = (g_1g_2,h_1,h_2)$ where $\displaystyle g_1g_2$ is the binary operation on $\displaystyle G$ and $\displaystyle h_1h_2$ is the binary operation on $\displaystyle H$ then $\displaystyle *$ defines a binary operation.

    Proving the condition for $\displaystyle G\times H$ to be a group are straightforward. For example, if you chose $\displaystyle (e,e')$ where $\displaystyle e$ is identity element in $\displaystyle G$ and $\displaystyle e'$ where $\displaystyle e'$ is identity element in $\displaystyle H$. Then $\displaystyle (g,h)*(e,e') = (ge,he')=(g,h)$. And similarly $\displaystyle (e,e')*(g,h) = (g,h)$. Try to prove the other properties.
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