9.4 Direct Product
In this section we’ll study a simple way of combining two groups to build a new, larger, group. We will do that by generalising the idea of the Cartesian product of two sets (Definition 6.1).
The direct product H\times K of groups is itself a group.
Associativity of H\times K follows from associativity of H and K, since if x,x',x''\in H and y,y',y''\in K, then \begin{align*} \left((x,y)(x',y')\right)(x'',y'')&=(xx',yy')(x'',y'')\\ &=(xx'x'',yy'y'')\\ &=(x,y)(x'x'',y'y'')\\ &=(x,y)\left((x',y')(x'',y'')\right). \end{align*}
If e_H and e_K are the identity elements of H and K, then (e_H,e_K) is the identity element of H\times K, since if x\in H and y\in K, then (e_H,e_K)(x,y)=(e_Hx,e_Ky)=(x,y)=(xe_H,ye_K)=(x,y)(e_H,e_K).
The inverse of (x,y)\in H\times K is (x^{-1},y^{-1}), since (x,y)(x^{-1},y^{-1})=(xx^{-1},yy^{-1})=(e_H,e_K) =(x^{-1}x,y^{-1}y)=(x^{-1},y^{-1})(x,y).□
Notice that for all aspects of the group structure, we simply apply the corresponding idea in the first coordinate for H and in the second coordinate for K. This is generally how we understand H\times K, by considering the two coordinates separately, and if we understand H and K, then H\times K is easy to understand. For example, it is easy to see that, for any i\in\mathbb{Z} and any (x,y)\in H\times K, (x,y)^i=(x^i,y^i), so also taking powers in a direct product is just a matter of taking powers of the coordinates separately.
Here are some very easy consequences of the definition.
Let H and K be (multiplicatively-written) groups, and let G=H\times K be the direct product.
G is finite if and only if both H and K are finite, in which case |G|=|H||K|.
G is abelian if and only if both H and K are abelian.
If G is cyclic then both H and K are cyclic.
We go through all three statements.
This is just a familiar property of Cartesian products of sets (that was left as an exercise).
Suppose H and K are abelian, and let (x,y),(x',y')\in G. Then (x,y)(x',y')=(xx',yy')=(x'x,y'y)=(x',y')(x,y), and so G is abelian.
Suppose G is abelian, and x,x'\in H, then (xx',e_K)=(x,e_K)(x',e_K)=(x',e_K)(x,e_K)=(x'x,e_K), and considering the first coordinates, xx'=x'x, and so H is abelian. Similarly K is abelian.
- Suppose G is cyclic, and (x,y) is a generator, so that every element of G is a power of (x,y). Let x'\in H. Then (x',e_K)\in G, so (x',e_K)=(x,y)^i=(x^i,y^i) for some i\in\mathbb{Z}, and so x'=x^i. So every element of H is a power of x, so H=\langle x\rangle is cyclic. Similarly K is cyclic.
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Let H and K be (multiplicatively-written) groups, and x\in H, y\in K elements with finite order. Then (x,y)\in H\times K has finite order equal to the lowest common multiple \operatorname{lcm}\left(\operatorname{ord}_H(x),\operatorname{ord}_K(y)\right).
Let i\in\mathbb{Z}. Then (x,y)^i=(e_H,e_K) if and only if x^i=e_H and y^i=e_K, which is the case if and only if i is divisible by both \operatorname{ord}_H(x) and by \operatorname{ord}_K(y). The least positive such i is \operatorname{lcm}\left(\operatorname{ord}_H(x),\operatorname{ord}_K(y)\right), and so this is the order of (x,y).
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We can now decide precisely when the direct product of cyclic groups is cyclic.
Let H=C_n and K=C_m be finite cyclic groups. Then C_n\times C_m is cyclic if and only if \gcd(n,m)=1.
Let (x,y)\in H\times K. Then \operatorname{ord}_H(x)\leq |H| and \operatorname{ord}_K(y)\leq |K|. So by Proposition 9.26, \operatorname{ord}_{H\times K}(x,y)=\operatorname{lcm}\left(\operatorname{ord}_H(x),\operatorname{ord}_K(y)\right) \leq\operatorname{ord}_H(x)\operatorname{ord}_K(y)\leq |H||K|, where the first inequality is an equality if and only if \operatorname{ord}_H(x) and \operatorname{ord}_K(y) are coprime, and the second inequality is an equality if and only if H=\langle x\rangle and K=\langle y\rangle.
Since |H\times K|=|H||K|, H\times K is cyclic if and only if it has an element of order |H||K|, which by the argument above is true if and only if \gcd\left(\operatorname{ord}_H(x),\operatorname{ord}_K(y)\right)=1.□
Since cyclic groups of the same order are isomorphic, the last theorem says that C_m\times C_n\cong C_{mn} if and only if \gcd(m,n)=1.
Felix Klein, German mathematician (1849 - 1925) applied ideas from group theory to geometry, which led to the emergence of transformation groups. These were some of the first example of infinite groups (beyond the groups (\mathbb{R},+), (\mathbb{R}\setminus\{0\},\times), which weren’t really studied as groups). Klein wrote about the Klein 4-group, although he called it Vierergruppe (meaning “four-group”), as it is the smallest group which is not cyclic.
C_2\times C_3\cong C_6
C_2\times C_9\cong C_{18}
C_{90}\cong C_2\times C_{45}\cong C_2\times (C_9\times C_5)
We can clearly extend the definition of a direct product of two groups to three, four or more groups and make sense of things like G\times H\times K, which would be a group whose elements are the ordered triples (x,y,z) with x\in G, y\in H and z\in K.
We can then write the last example as C_{90}\cong C_2\times C_9\times C_5.
More generally, if n=p_1^{r_1}p_2^{r_2}\dots p_k^{r_k}, where p_1,p_2,\dots,p_k are distinct primes, then C_n\cong C_{p_1^{r_1}}\times C_{p_2^{r_2}}\times\dots\times C_{p_k^{r_k}}, so that every finite cyclic group is isomorphic to a direct product of groups with order powers of primes.
We’ll finish this section with a different example of a familiar group that is isomorphic to a direct product.
(\mathbb{R}\setminus\{0\},\times)\cong(\mathbb{R}_{>0},\times)\times\left(\{1,-1\},\times\right).
Define a map \varphi:\mathbb{R}_{>0}\times\{1,-1\}\to\mathbb{R}\setminus\{0\} by \varphi(x,\varepsilon)=\varepsilon x. This is clearly a bijection, and \varphi\left((x,\varepsilon)(x'\varepsilon')\right) =\varphi(xx',\varepsilon\varepsilon')=\varepsilon\varepsilon'xx' =(\varepsilon x)(\varepsilon'x')=\varphi(x,\varepsilon)\varphi(x',\varepsilon'), so that \varphi is an isomorphism.
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