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 \[(x,y)(x^{-1},y^{-1})=(xx^{-1},yy^{-1})=(e_H,e_K) =(x^{-1}x,y^{-1}y)=(x^{-1},y^{-1})(x,y).\]

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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.

\(G\) 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.

Proof.

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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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 \((x,y)\).

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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 \(\gcd\left(\operatorname{ord}_H(x),\operatorname{ord}_K(y)\right)=1\).

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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.

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