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The question is as follows:

Let $G_1$ and $G_2$ be groups. Define $\pi_1 : G_1 \times G_2 \rightarrow G_1$ by $\pi_1((a_1,a_2))=a_1.$ Define $\pi_2 : G_1 \times G_2 \rightarrow G_2$ by $\pi_2((a_1,a_2))=a_2.$

Let $G$ be any group, and let $\phi : G \rightarrow G_1 \times G_2$ be a function. Show that $\phi$ is a group homomorphism if and only if $\pi_1\circ \phi$ and $\pi_2\circ \phi$ are both group homomorphisms.

Going forward is easy, but I'm not sure how to do the backwards direction (proving that if $\pi_1\circ \phi$ and $\pi_2\circ \phi$ are both group homomorphisms, then $\phi$ is a group homomorphism.)

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  • $\begingroup$ You need to use the universal property of the direct product. What could the unique morphism from $G$ to $G_1\times G_2$ be? $\endgroup$ – Robert Wolfe Mar 1 '14 at 22:29
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It's also easy $$\varphi(gg')=((\pi_1\circ\varphi)(gg'),(\pi_2\circ\varphi)(gg'))={((\pi_1\circ\varphi)(g)\cdot(\pi_1\circ\varphi)(g')),(\pi_2\circ\varphi)(g)\cdot(\pi_2\circ\varphi)(g')))}=((\pi_1\circ\varphi)(g),(\pi_2\circ\varphi)(g))\cdot((\pi_1\circ\varphi)(g'),(\pi_2\circ\varphi)(g'))=\varphi(g)\varphi(g')$$

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Look at it this way: $\phi: G \to G_1\times G_2$ takes an element of $G$ and returns a pair: one elment from $G_1$, and one element from $G_2$, i.e. $\phi(g)=(g_1,g_2)$.

Can you manipulate this expression to get equations of the form $g_1=?$, $g_2=?$ The goal is to express $\phi$ entirely in terms of $\pi_1\circ\phi$ and $\pi_2\circ\phi$, which will allow you to explicitly calculate $\phi(gh)$ and $\phi(g)\phi(h)$.

As a side note, I wish that it were not so common to post questions without context. Math is a very personal thing, and it's almost impossible to give good answers when you don't know where somebody is coming from, what they know, why they're asking a question, or what they hope to get out of it. I'm not saying that every question should come with an autobiography, but a single sentence like "I am working through Dummit and Foote independently," or "I'm stuck on my homework," or even "I found this fun problem," really makes for a healthier interaction in my opinion.

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  • $\begingroup$ This was an old homework question that many students in my class got wrong because we assumed $\pi_1$ and $\pi_2$ were one-to-one. $\endgroup$ – Schala Mar 1 '14 at 22:50
  • $\begingroup$ Also, this is my first abstract algebra class, but I think I've seen more stuff on the direct product in a proof-writing class I took last semester. $\endgroup$ – Schala Mar 1 '14 at 22:53
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Let $ \phi(g) = (g_1, g_2)$ for each $g \in G$. ,and consider $x, y$ in $G$. We need to show: $\phi(xy) = \phi(x) \phi(y)$.

Put $\phi(x) = (x_1, y_1)$, and $\phi(y) = (x_2, y_2)$. Then $\phi(x) \phi(y) = (x_1x_2, y_1y_2)$.

Next we compute $\phi(xy)$. Let's put $\phi(xy) = (a,b)$.

Then $$\pi_1(\phi(xy)) = a,\ \ and\ \ \pi_2(\phi(x*y)) = b$$. But$$\pi_1(\phi(xy)) = \pi_1(\phi (x))\pi_1(\phi(y)) \implies a = x_1x_2$$, and similarly $$\pi_2(\phi(xy)) = \pi_2(\phi(x))\pi_2(\phi(y)) \implies b = y_1y_2$$.

So $(a,b) = (x_1x_2, y_1y_2) \implies \phi(xy) = \phi(x)\phi(y) \implies \phi$ is a homomorphism.

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  • $\begingroup$ Please, learn to use $$, this is unreadable. $\endgroup$ – user2345215 Mar 1 '14 at 23:19

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