# Direct Product

Printable View

• Jan 12th 2008, 11:28 AM
smoothman
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..
• Jan 12th 2008, 02:40 PM
ThePerfectHacker
Quote:

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 $*$ such that $*:G\times G\mapsto G$, so of course it is closed. However, a subgroup needs to satisfy the closure property.

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

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