The Universal Property of the Direct Product in Groups

[Deveno]

Motivation​

With groups, one often seeks to create larger groups out of smaller ones, or conversely to break down large groups into easier-to-understand pieces. One construction commonly used for this purpose is the direct product.

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The Usual Definition​

The direct product of two groups $(G,\ast)$ and $(H,\ast')$ is defined to be the set

$G \times H = \{(g,h): g \in G, h \in H\}$

together with the binary operation

$(g,h)\star(g',h') = (g\ast g',h\ast' h')$ for all $g,g' \in G$ and $h,h' \in H$.

One then shows that $\star$ is associative, that $(e_G,e_H)$ is an identity for this group, and that for each $(g,h)$, the element $(g^{-1},h^{-1})$ is an inverse.

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An Alternative Definition​

I want to discuss another definition and show that it is "essentially the same" as the one above. This may seem a bit strange at first, so I will try to illuminate it as we go along.

We define a direct product $P$ of the two groups $G,H$ to be:

1. A group $P$, along with:
2. Two group homomorphisms $p_1:P \to G$ and $p_2: P \to H$ such that:
3. If $K$ is ANY other group, along with ANY pair of group homomorphisms $f_1:K \to G$ and $f_2:K \to H$, there is a UNIQUE homomorphism $\phi:K \to P$ so that: $p_1 \circ \phi = f_1$ and $p_2 \circ \phi = f_2$.

Some initial comments: first, note I said "a" direct product, not "the" direct product. It is thus conceivable we might have more than one, or perhaps none at all. It is not clear from 1–3 that any group whatsoever satisfies these properties, or that even if one does, it is unique. We shall see that the former is true, and the latter not quite so true.

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Two Non-Obvious Consequences​

Before we go any further, I will state two assertions about $P$ that are not obvious at all from 1–3.

Lemma 1: there exist unique distinguished injections $G \to P$ and $H \to P$.

Lemma 2: $p_1,p_2$ are surjective.

Both of these assertions will follow from the following basic fact about functions:

If $f \circ g$ is bijective, then $f$ is surjective, and $g$ is injective. If you do not believe this, I urge you to prove it on your own (it's not that hard, really).

To see this, suppose $K = G$, with $f_1: G \to G$ the identity map, and $f_2: G \to H$ the trivial map that sends everything to $e_H$. From our definition of $P$ (if it exists), we get a unique homomorphism

$\phi_1:G \to P$

with

$p_1 \circ \phi_1 = 1_G$.

Now the identity map on $G$ is clearly a bijection, which shows that $\phi_1$ is the unique distinguished injection claimed in Lemma 1, and that $p_1$ is a surjection, as claimed in Lemma 2.

We can do a similar thing with $H$.

Note also that, from what we've defined so far,

$p_2 \circ \phi_1 = 0$

(using $0$ to mean $0(g) = e_H$ for all $g \in G$), that is,

$\text{im }\phi_1 \subseteq \text{ker }p_2$.

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Uniqueness Up to Isomorphism​

Our next result is as good as we can get regarding uniqueness of a direct product:

Theorem 1: If $P,P'$ are two direct products of $G$ and $H$, then $P \cong P'$.

To see this, suppose first that $P$ is a direct product, and let $K = P'$, with the two maps

$p_1':P' \to G$
$p_2':P' \to H$

that $P'$ comes with (also being a direct product).

We thus have a unique homomorphism $\psi: P' \to P$ (with, of course, $p_1 \circ \psi = p_1'$ and $p_2 \circ \psi = p_2'$).

Exchanging the places of $P$ and $P'$ we also have a unique homomorphism $\theta: P \to P'$ (and with $p_1' \circ \theta = p_1$ and $p_2' \circ \theta = p_2$).

Composing the two gives a homomorphism $\psi \circ \theta: P \to P$. Note that we have:

$p_1 \circ \psi \circ \theta = p_1' \circ \theta = p_1$
$p_2 \circ \psi \circ \theta = p_2' \circ \theta = p_2$

Now, consider $K = P$ along with the two maps:

$f_1: P \to G$ given by: $f_1 = p_1$
$f_2: P \to G$ given by: $f_1 = p_2$

We have a UNIQUE homomorphism $\phi: P \to P$ such that $p_1 \circ \phi = p_1$ and $p_2 \circ \phi = p_2$.

Since the identity homomorphism makes this true, it must be (by uniqueness) that $\phi = 1_P$.

Since this is also true of the homomorphism $\psi \circ \theta$, it must be true that $\psi \circ \theta = 1_P$.

A similar argument shows $\theta \circ \psi = 1_{P'}$, so these are both bijective homomorphisms, that is: isomorphisms.

Now, not only does this establish that any two direct products of $G$ and $H$ are isomorphic, but also: there is a uniquely defined isomorphism between them.

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Recovering $G \times H$ from the Universal Property​

Theorem 2: $G \times H$ (our "old" definition of direct product) is a direct product of $G$ and $H$ (under our "new" definition).

Clearly, condition 1 is satisfied, but we need to find a pair of group homomorphisms to serve as our $p_1,p_2$. Let:

$\pi_1: G\times H \to G$ be defined by: $\pi_1(g,h) = g$
$\pi_2: G\times H \to H$ be defined by: $\pi_2(g,h) = h$.

Now we must show that given any other group $K$ along with two homomorphisms

$f_1: K \to G$
$f_2: K \to H$

we can find a UNIQUE homomorphism $\phi:K \to G \times H$ with

$\pi_1 \circ \phi = f_1$
$\pi_2 \circ \phi = f_2$.

Let us determine what $\phi$ must be (if it exists). Suppose $\phi(k) = (g_k,h_k)$.

Then $g_k = \pi_1(g_k,h_k) = \pi_2(\phi(k)) = (\pi_1 \circ \phi)(k) = f_1(k)$, and similarly we must have $h_k = f_2(k)$.

So, as a FUNCTION, we must have: $\phi(k) = (f_1(k),f_2(k))$.

Is this a homomorphism?

$\phi(kk') = (f_1(kk'),f_2(kk')) = (f_1(k)f_1(k'), f_2(k)f_2(k')) = (f_1(k),f_2(k))(f_1(k'),f_2(k')) = \phi(k)\phi(k')$.

So, yes, yes it is.

Thus this $\phi$ is clearly the ONLY homomorphism we could have, and it works (yay!). Now we can establish:

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A Quotient Description​

Theorem 3: For a direct product $P$ of $G$ and $H$, and the unique monomorphism $\phi_1: G \to P$:

$H \cong P/\phi_1(G)$ (a similar statement holds for $G$ and $H$ swapping places).

Consider the isomorphism $\psi:G\times H \to P$.

We have the uniquely defined injection

$i_1: G \to G \times H$ given by $g \mapsto (g,e_H)$.

Since, for any $(a,b) \in G \times H$, we have

$(a,b)(g,e_H)(a,b)^{-1} = (a,b)(g,e_H)(a^{-1},b^{-1}) = (aga^{-1},e_H)$

it follows that the image under $i_1$ of $G$ in $G \times H$ (namely: $G \times \{e_H\}$) is a normal subgroup of $G \times H$.

Thus $\psi(G \times \{e_H\}) = \phi_1(G)$ is a normal subgroup of $\psi(G \times H) = P$.

As we saw before, it is clear that $\text{im }i_1 \subseteq \text{ker }\pi_2$.

Now however, we see as well that if $(g,h) \in \text{ker }\pi_2$, then $h = e_H$, and thus

$(g,h) \in G \times \{e_H\} = \text{im }i_1$.

This means $\text{ker }\pi_2 = \text{im }i_1$, from whence we see that:

$H \cong (G \times H)/\text{ker }\pi_2 = (G \times H)/\text{im }i_1 \cong \psi(G \times H)/\psi(\text{im }i_1) = P/\phi_1(G)$.

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Short Exact Sequences​

In summary, we have a short exact sequence:

$1 \to G \stackrel{\phi_1}{\to}P \stackrel{p_2}{\to}H \to 1$

which is left-split: because we have $p_1:P \to G$ such that $p_1 \circ \phi_1 = 1_G$.

This is actually a THIRD way to characterize direct products. Clearly, any direct product generates a left-split short exact sequence. Surprisingly, the converse is true:

Theorem 4:

If $1 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 1$ is a left-split short exact sequence, then $B$ is a direct product of $A$ and $C$.

Since the sequence is left-split, we have a homomorphism $h: B \to A$ such that $h \circ f = 1_A$.

Define $\theta: B \to A\times C$ by:

$\theta(b) = (h(b),g(b))$ (often this map is written $h \times g$).

This is a homomorphism because $h$ and $g$ are. Suppose $\theta(b) = (e_A,e_C)$. By exactness, we have: $b = f(a)$, for some $a \in A$.

Now $e_A = h(b) = h(f(a)) = (h \circ f)(a) = 1_A(a) = a$, so $b = f(e_A) = e_B$, thus $\theta$ is injective.

Now suppose that $(a,c)$ is any element of $A \times C$. Since $g$ is surjective, we can find $b \in B$ such that $g(b) = c$.

Note that $g^{-1}(\{c\}) = \{bf(a): a \in A\}$. Now if we take $x = [h(b)]^{-1}a \in A$, then:

$\theta(bf(x)) = (h(bf(x)),g(bf(x))) = (h(b)h(f(x)),g(b)g(f(x))) = (h(b)(h \circ f)(x), g(b)e_C) = (h(b)x,g(b)) = (h(b)[h(b)]^{-1}a,c) = (e_Aa,c) = (a,c)$,

so $\theta$ is surjective.

Note that $(\theta \circ f)(a) = (h(f(a)),g(f(a))) = (a,e_C)$ and $(\pi_2 \circ \theta)(b) = g(b)$

so that $\theta$ transforms $f$ and $g$ into the "standard" injection of $A$ into $A \times C$ and projection of $A \times C$ onto $C$.

 

[Greg Bernhardt]

Thanks @Deveno! What forum can we move this to?