How to proof that external direct product of groups is itself a group?

Kurchi

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How to proof that external direct product of groups is itself a group? I've found a website which contains the proof -https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group
Please read this proof from this website . My problem is, in this proof in the given link has a binary composition 'o' for
$G_1$
*
$G_2$
. But binary composition for
$G_1$
and
$G_2$
are
$o_1$
and
$o_2$
respectively. What is 'o' and where did it came from?
 
How to proof that external direct product of groups is itself a group? I've found a website which contains the proof -https://proofwiki.org/wiki/External_Direct_Product_of_Groups_is_Group
Please read this proof from this website . My problem is, in this proof in the given link has a binary composition 'o' for
$G_1$
*
$G_2$
. But binary composition for
$G_1$
and
$G_2$
are
$o_1$
and
$o_2$
respectively. What is 'o' and where did it came from?
Allow me to change the notation: Suppose that \((H,\circ_1)~\&~(K,\circ_2)\) are groups then their direct product is
\(H\times K=\{(h,k) : h\in H~\&~k\in K\}\) with operation defined as \((h,k)\circ (m,n)=(h\circ_1 m, k\circ_2 n)\).
It is a standard exercise to show that as defined \((H\times K,\circ)\) is a group.
Show that if \(e_1\) is the identity of \(H\) and \(e_2\) is the identity of \(K\) then show that \((e_1,e_2)\) is the identity of \(H\times K\).
If \((a,b)\in H\times K\) what would the inverse element be?
As defined, can you prove that the operation \({\large\circ}\) is associative?
 
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