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Direct Product of GroupsLet G an H be groups. The Cartesian product
(g,h)(g',h') = (gg',hh').
It is clear that (1) defines a binary operation on
(g,h)(1_{G},1_{H}) = (g·1_{g}, h·1_{H}) = (g,h),
(1_{G},1_{H})(g,h) = (1_{G}·g,1_{H}·h) = (g,h),
we see that (1_{G},1_{H}) is an identity element. Moreover
(g,h)(g^{1},h^{1}) = (g^{1},h^{1})(g,h) = (1_{G},1_{H}),
so that every element of
Definition 2: Example 1: Let
= {(0,0),(0,1),(1,0),(1,1)}.
The identity element is
AB = BA, A^{2} = B^{2} = I, C = AB.
Thus Example 2: Let
= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}.
Note that (0,0) is the identity and that if
^{2} = (0,2),
^{3} = (1,0),
^{4} = (0,1),
^{5} = (1,2),
^{6} = (0,0),
so that has order 6. Thus, Let G_{1},...,G_{n} be a finite collection of groups. The direct product G_{1} ... G_{n} is the group whose elements are the ordered ntuples (g_{1},....g_{n}), where
(g_{1},...,g_{n})(g_{1}',...,g_{n}') = (g_{1}g_{1}',...,g_{n}g_{n}').
The verification that If G_{1},G_{2}, and G_{3} are groups, we can form the direct product
:(G_{1} G_{2}) G_{3} G_{1} G_{2} G_{3},
(((g_{1},g_{2}),g_{3})) = (g_{1},g_{2},g_{3}),
is a surjective isomorphism, so that
(G_{1} G_{2}) G_{3} G_{1} G_{2} G_{3}.
Similarly,
G_{1} (G_{2} G_{3}) G_{1} G_{2} G_{3}.
Subsequently, we will always identify the groups (G_{1} G_{2}) G_{3}, G_{1} (G_{2} G_{3}), G_{1} G_{2} G_{3}. Similar comments apply to direct products of more than three groups. For example, G_{1} (G_{2} G_{3}) G_{4} will be identified with G_{1} G_{2} G_{3} G_{4}, and so on. The next result is a convenient one for decomposing a given group into a direct product of two subgroups.
Theorem 3: Let G be a group. Let (1) H K = {1}. (2) If h H, k K, then (3) If g G, then there exist
Then Proof: First let us prove that if
h'^{1}h = k'k^{1} H K = {1},
h'^{1}h = k'k^{1} = 1,
h = h', k = k'.
Let
(gg') = (hh',kk')
= (h,k)(h',k')
= (g)(g')
Thus, is a homomorphism. A simple, but important application of Theorem 3 is the following.
Theorem 4: Let m and n be positive integers such that Proof: Let a be a generator of G. By Theorem 16 of the section of subgroups, the order of a^{m} is n and the order of a^{n} is m. Let
Therefore, it suffices to prove that
g = a^{r} = a^{mcr+ndr}
= (a^{m})^{cr}·(a^{n})^{dr}.
Therefore (1)(3) of Theorem 3 hold, and Corollary 5: Let G be a cyclic group of order p_{1}^{a1}p_{2}^{a2}...p_{t}^{at}, where p_{1},...,p_{t} are distinct primes and a_{1},...,a_{t} are positive integers. Then Proof: Let us proceed by induction on t. The assertion is true for 
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