1. ## 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..

2. Originally Posted by smoothman
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.