Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Direct Product of Groups

Point of post: In this post we cover the concept of direct product of finitely many groups


By my hand being forced, I had to go quickly through all of this group theory. Consequently, things were done out of order and there is a particular dearth of examples in my posts. This post will be the first of two (the second will be on semidirect products) posts where we describe an interesting way to combine two (or more) groups to create a third. The concept is similar to most other mathematical structures. In essence, there is a canonical way to define a group structure on the Cartesian product of a finite number of groups. In fact, there is a canonical way to define the direct product and direct sum (the two coincide when only finitely many groups are involved) to arbitrary collections of groups. Sadly, time does not permit this level of detail

Direct Product of Groups

Let G_1,\cdots,G_n be finitely many groups. Then, define the an operation on the Cartesian product\displaystyle \prod_{j=1}^{m}G_j by


where the operations in each coordinate given in the above is the given operation in each of the groups G_1,\cdots,G_n. It’s easy to verify that this defines a group structure with identity element (e_{G_1},\cdots,e_{G_n}) and inverses given by


we call this group the direct product of G_1,\cdots,G_n and denote it \displaystyle \prod_{j=1}^{n}G_j.  Since we will often have need to use it for each k\in[n] we denote the set

\{(e,\cdots,\underbrace{g}_{k^{\text{th}}\text{ coordinate}},\cdots,e):g\in G_k\}

by \tilde{G_k}

We first notice that the direct product of groups is canonically equipped with two homomorphisms. Namely

\displaystyle \pi_k:\prod_{j=1}^{m}G_j\to G_k:(g_1,\cdots,g_k,\cdots,g_n)\mapsto g_k

Indeed this is a homomorphism since

\displaystyle \begin{aligned}\pi_k\left((g_1,\cdots,g_k,\cdots,g_n)(g'_1,\cdots,g'_k,\cdots,g'_n)\right) &=g_kg'_k\\ &=\pi_k\left((g_1,\cdots,g_k,\cdots,g_n)\right)\pi_k\left((g'_1,\cdots,g'_k,\cdots,g'_n)\right)\end{aligned}

and since evidently

\ker\pi_k=G_1\times\cdots\times\underbrace{\{e\}}_{k^{\text{th}}\text{ coordinate}}\times\cdots\times G_k

and \pi_k is surjective  we may conclude that \displaystyle \tilde{G_k}\unlhd \prod_{j=1}^{n}G_j for each k\in[n] and by the First Isomorphism Theorem that

\displaystyle \left(\prod_{j=1}^{n}G_j\right)/\left(G_1\times\cdots\{e\}\times\cdots\times G_n\right)\cong G_k

This mapping is called the projection of \displaystyle \prod_{j=1}^{n}G_k onto its k^{\text{th}} coordinate.

In the other direction we have the k^{\text{th}} canonical injection, denoted \iota_k, given by

\displaystyle \iota_k:G_k\to \prod_{j=1}^{n}G_j:g\mapsto (e,\cdots,\underbrace{g}_{k^{\text{th}}\text{ coordinate}},\cdots,e)

evidently this is a monomorphism and \iota_k\left(G_k\right)=\tilde{G_k} from where it follows that G_k\cong \tilde{G_k}. Our first result are certain universal properties about homomorphisms in and out of a product space. In particular:

Theorem: Let G,G_1,\cdots,G_n be groups and \phi_k\in\text{Hom}\left(G,G_k\right) for k\in[n]. Then there exists a unique homomorphisms

\displaystyle \phi:G\to\prod_{j=1}^{n}G_j

such that \pi_k\circ\phi=\phi_k for each k\in[n]. In particular


Proof: It’s evident that \theta is any mapping which satisfies the above properties then one must have that


and thus it remains to show that this is indeed a homomorphism. But, this follows since

\displaystyle \begin{aligned}\phi\left((g_1,\cdots,g_n)(g'_1,\cdots,g'_n)\right) &= \phi\left((g_1g'_1,\cdots,g_ng'_n)\right)\\ &= \left(\phi_1(g_1g'_1),\cdots,\phi_n(g_ng'_n)\right)\\ &=\left(\phi_1(g_1)\phi_1(g'_1),\cdots,\phi_n(g_n)\phi_n(g'_n)\right)\\ &= \left(\phi_1(g_1),\cdots,\phi_n(g_n)\right)\left(\phi_1(g'_1),\cdots,\phi_n(g'_n)\right)\\ &= \phi\left((g_1,\cdots,g_n)\right)\phi\left((g'_1,\cdots,g'_n)\right)\end{aligned}


In a dual manner there is a universal property for homomorphisms out of a product space. Namely:

Theorem: Let G,G_1,\cdots,G_n be groups and \phi_k\in\text{Hom}\left(G_k,G\right) be such that \phi_k\phi_\ell=\phi_\ell\phi_k (they commute in terms of multiplication, not in terms of composition) whenever \ell\ne k. Then, there is a unique homomorphism

\displaystyle \phi:\prod_{j=1}^{n}G_j\to G

such that \phi_k=\phi\circ \iota_k for each k\in[n]. In particular

\displaystyle \phi\left((g_1,\cdots,g_n)\right)=\phi_1(g_1)\cdots\phi_n(g_n)

Proof: It’s evident that any such map would have to be of the form

\displaystyle \phi\left((g_1,\cdots,g_n)\right)=\phi_1(g_1)\cdots\phi_n(g_n)


\displaystyle \begin{aligned}\phi\left((g_1,\cdots,g_n)\right) &= \phi\left(\iota_1(g_1)\cdots\iota_n(g_n)\right)\\ &= \phi\left(\iota_1(g_1)\right)\cdots\phi\left(\iota_n(g_n)\right)\\ &= \phi_1(g_1)\cdots\phi_n(g_n)\end{aligned}

So, it suffices to prove that this map is, in fact, a homomorphism. To do this we merely  note that

\displaystyle \begin{aligned}\phi\left((g_1,\cdots,g_n)(g'_1,\cdots,g'_n)\right) &= \phi\left((g_1g'_1,\cdots,g_ng'_n)\right)\\ &= \phi_1(g_1g'_1)\cdots\phi_n(g_ng'_n)\\ &= \phi_1(g_1)\phi_1(g'_1)\cdots\phi_n(g_n)\phi_n(g'_n)\\ &= \phi_1(g_1)\cdots\phi_n(g_n)\phi_1(g'_1)\cdots\phi_n(g'_n)\\&= \phi\left((g_1,\cdots,g_n)\right)\phi\left((g'_1,\cdots,g'_n)\right)\end{aligned}

from where the conclusion follows. \blacksquare


1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200


January 10, 2011 - Posted by | Algebra, Group Theory | ,


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