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).

Definition 9.23:
Let H and K be (multiplicatively-written) groups. The direct product H\times K of H and K is the Cartesian product of the sets H and K, with the binary operation (x,y)(x',y')=(xx',yy') for x,x'\in H and y,y'\in K.
Proposition 9.24:

The direct product H\times K of groups is itself a group.

Proof.

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.

Proposition 9.25:

Let H and K be (multiplicatively-written) groups, and let G=H\times K be the direct product.

  1. G is finite if and only if both H and K are finite, in which case |G|=|H||K|.

  2. G is abelian if and only if both H and K are abelian.

  3. If G is cyclic then both H and K are cyclic.

Proof.

We go through all three statements.

  1. This is just a familiar property of Cartesian products of sets (that was left as an exercise).

  2. 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.

  1. 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.

Remark:
The converse of c.) is not true in general: the direct product of cyclic groups may not be cyclic. For example, if H=K=C_2, then (x,y)^2=(e_H,e_K) for every (x,y)\in C_2\times C_2, so C_2\times C_2 has no element of order |H\times K|=4, and so can’t be cyclic.
Proposition 9.26:

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).

Proof.

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).

We can now decide precisely when the direct product of cyclic groups is cyclic.

Theorem 9.27:

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.

Proof.

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.

Remark:
While the above proof tells us the existence of an isomorphism between C_m\times C_n and C_{mn} it does not tell us explicitly what the isomorphism is. However, we can use Theorem 10.14 to construct the isomorphism between C_m \times C_n and C_{mn}. Let C_m = \langle x \rangle, C_n = \langle y \rangle , C_{mn} = \langle z \rangle and define \varphi:C_m\times C_n \to C_{mn} by \varphi((x^i,y^j)) = z^k where by Theorem 10.14 k is such that k\equiv i \pmod{m} and k\equiv j \pmod{n}.
Example:
C_2\times C_2 and C_4\times C_2 are not cyclic.
Remark:
C_2\times C_2 is an abelian group of order 4 such that every element apart from the identity has order 2. It is easy to check that if G=\{e,a,b,c\} is any group with these properties, then ab=c=ba, ac=b=ca and bc=a=cb: i.e., the product of any two of the three non-identity elements is the other non-identity element. This means that there is only one possible multiplication table for such a group, and so any two groups with these properties are isomorphic.
Definition 9.28:
A Klein 4-group is a group of order 4 such that every element except the identity has order 2.
History:

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.

Example:
  • 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.

Proposition 9.29:

(\mathbb{R}\setminus\{0\},\times)\cong(\mathbb{R}_{>0},\times)\times\left(\{1,-1\},\times\right).

Proof.

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.