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- Feb 15, 2012

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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 goes to show that $\star$ is associative, that $(e_G,e_H)$ is an identity for this group and that for each $(g,h)$ that $(g^{-1},h^{-1})$ is an inverse.

I want to discuss another way, and show that it is "essentially the same" as the above. This may seem a bit strange at first, so I will try to illuminate it as best I can, 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 $p_1 \to G$ 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.

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

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

__Lemma 1__**: $p_1,p_2$ are surjective.**

__Lemma 2__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 see that we have 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 we claimed existed in lemma 1, and that $p_1$ is a surjection, as claimed in lemma 2.

We can do a similar thing with $H$.

Note that we have, just from what we've defined so far, something else:

$p_2 \circ \phi_1 = 0$ (using "$0$" to mean $0(g) = e_H$, for all $g \in G$), that is:

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

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

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

__Theorem 1__To see this, suppose first that $P$ is a direct product, and let $K = P'$, with the two maps:

$p_1'' \to G$

$p_2': P' \to H$

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.

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

__Theorem 2__Clearly, condition 1 is satisfied, but we need to find a pair of group homomorphisms to serve as our $p_1,p_2$. However, we don't have to look far, 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 have to 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's focus on what $\phi$ would have to be, first (if it indeed 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, just 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.

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

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

__Theorem 3__$H \cong P/\phi_1(G)$ (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 $(g,h) \in \text{ker }\pi_2$, that $h = e_H$, and thus that:

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

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)$.

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 $p_1 \to G$ 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 $C$.