Let \((G,*)\) and \((H,\circ)\) be two groups and \(G\times H\) be the
set of ordered pairs \((g,h)\) with \(g\in G\) and \(h\in H.\) A multiplication
on \(G\times H\) is defined by
\[
(g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2})
\]
for all \(g_{1},g_{2}\in G\) and \(h_{1},h_{2}\in H\).
Show that, with this multiplication, \(G\times H\) is a group.
State whether the following are true or false and prove your answers.
\(G\times H\) is abelian if and only if both \(G\) and \(H\) are abelian.
\(G\times H\) contains a subgroup isomorphic to \(G\).
\(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\) is isomorphic to \(\mathbb{Z}_{4}.\)
\(S_{2}\times S_{3}\) is isomorphic to \(S_{6}.\)
{[}\(\mathbb{Z}_{n}\) is the cyclic group of order \(n\), and \(S_{n}\)
is the permutation group on \(n\) objects.{]}
Solution: Claim: \(G \times H\) is a group. (Called the product group).
Proof: Checking the group axioms:
(Closure) is inherited from \(G\) and \(H\), since \(g_1 * g_2 \in G\) and \(h_1 \circ h_2 \in H\)
(Inverses) If \((g,h) \in G \times H\) then consider \((g^{-1}, h^{-1})\) and we have \((g^{-1}, h^{-1})(g,h) = (e_G,e_H) = (g,h)(g^{-1}, h^{-1})\)
Claim: \(G \times H\) is abelian iff \(G\) and \(H\) are.
Proof: \(\Rightarrow\) Suppose \(g_1, g_2 \in G\) and \(h_1, h_2 \in H\) then \((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (h_1,h_2)(g_1, g_2) = (g_2 * g_1, h_2 \circ h_1)\) so \(g_1*g_2 = g_2*g_1\) and \(h_1 \circ h_2 = h_2 \circ h_1\), therefore \(G\) and \(H\) are commutative.
\(\Leftarrow\) If \(H\) and \(G\) are commutative then:
\((g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (g_2 * g_1, h_2 \circ h_1) = (h_1,h_2)(g_1, g_2)\) so \(G \times H\) is commutative.
Claim: \(G\times H\) contains a subgroup isomorphic to \(G\). Consider the subset \(S = \{(g,e_H) : g \in G \}\). Then this is a subgroup isomorphic to \(G\) with isomorphism given by \(\phi : S \to G\) by \(\phi((g,e_H)) = g\)
If \(x \in \mathbb{Z}_2 \times \mathbb{Z}_2\) then \(x^2 = e\), but \(1\) does not have order 2 in \(\mathbb{Z}_4\)
\(S_2 \times S_3\) has order \(2 \times 6 = 12\). \(S_6\) has order \(6! \neq 12\)