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Group theory: epimorphisms

TheBigBadBen

Active member
May 12, 2013
84
Interesting question I've happened upon:

If there is an epimorphism (i.e. onto homomorphism) $\phi:G\times G \to H\times H$, is there necessarily an epimorphism $\psi:G\to H$? If not, under what conditions can we ascertain such an epimorphism given the existence of $\phi$?

I would think that this is true for abelian groups $G$ and $H$ by their respective decompositions, but I'm not sure how I'd show that to be the case. Also, I don't have a good guess for the general case. It seems clear that the isomorphism theorem should play a role at some point, but that doesn't lead to any obvious conclusions.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Interesting question I've happened upon:

If there is an epimorphism (i.e. onto homomorphism) $\phi:G\times G \to H\times H$, is there necessarily an epimorphism $\psi:G\to H$? If not, under what conditions can we ascertain such an epimorphism given the existence of $\phi$?

I would think that this is true for abelian groups $G$ and $H$ by their respective decompositions, but I'm not sure how I'd show that to be the case. Also, I don't have a good guess for the general case. It seems clear that the isomorphism theorem should play a role at some point, but that doesn't lead to any obvious conclusions.
Wouldn't there have to be? Let $h_{1},h_{2}\in H$ be arbitrary. Then there exist $g_{1},g_{2}\in G$ such that $\phi(g_{1},g_{2})=(h_{1},h_{2})$. Let $\psi$ be defined by the action of $\phi$ on only the first component. I think it'd be fairly straightforward to show that $\psi$ inherits all the required properties straight from $\phi$. Onto seems clear: since $h_{1}$ is arbitrary, and $\phi(g_{1},g_{2})=(h_{1},h_{2})$, it follows that $\psi(g_{1})=h_{1}$, and you have your element in $G$ that maps to $h_{1}$. And the fact that $\psi$ is a homomorphism follows from the fact that $\phi$ is, right?
 

TheBigBadBen

Active member
May 12, 2013
84
Wouldn't there have to be? Let $h_{1},h_{2}\in H$ be arbitrary. Then there exist $g_{1},g_{2}\in G$ such that $\phi(g_{1},g_{2})=(h_{1},h_{2})$. Let $\psi$ be defined by the action of $\phi$ on only the first component. I think it'd be fairly straightforward to show that $\psi$ inherits all the required properties straight from $\phi$. Onto seems clear: since $h_{1}$ is arbitrary, and $\phi(g_{1},g_{2})=(h_{1},h_{2})$, it follows that $\psi(g_{1})=h_{1}$, and you have your element in $G$ that maps to $h_{1}$. And the fact that $\psi$ is a homomorphism follows from the fact that $\phi$ is, right?
Your statement assumes that $\phi$ must act on the components separately. I thought this might be at first, but I don't believe this is necessarily the case.

A possible counter-example: $\phi:\mathbb {Z\times Z} \to \mathbb {Z}\times \mathbb Z$ given by $\phi(x,y) = (x+y,x-y)$ is an epimorphism in which you can't separate the components in the manner you described.

EDIT: The counter-example is not onto. Still trying to think of something to prove my point.
 
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TheBigBadBen

Active member
May 12, 2013
84

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Your statement assumes that $\phi$ must act on the components separately. I thought this might be at first, but I don't believe this is necessarily the case.

A possible counter-example: $\phi:\mathbb {Z\times Z} \to \mathbb {Z}\times \mathbb Z$ given by $\phi(x,y) = (x+y,x-y)$ is an epimorphism in which you can't separate the components in the manner you described.

EDIT: The counter-example is not onto. Still trying to think of something to prove my point.
I'm not seeing how the counter-example fails to be onto. Can you exhibit an element of $\mathbb{Z}$ that cannot be reached by the quantity $x+y$?
 

TheBigBadBen

Active member
May 12, 2013
84
I'm not seeing how the counter-example fails to be onto. Can you exhibit an element of $\mathbb{Z}$ that cannot be reached by the quantity $x+y$?
No, but I can certainly exhibit an element of $\mathbb{Z}\times\mathbb{Z}$ that cannot be reached by $(x+y,x-y)$: one example is $(0,1)$. In general, $x+y$ is even iff $x-y$ is even for $(x,y)\in\mathbb{Z\times Z}$.
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Ah. I think I've found the problem with my proposal. The problem with my example isn't surjectivity: it's whether it's well-defined. You could have $\phi(0,0)=(0,0)$ and $\phi(1,0)=(0,1)$. Then my $\psi$ function isn't well-defined. Hmm. I wonder if there's a way to fix my example. I was thinking maybe you could use equivalency classes, but then would you really be mapping from $\mathbb{Z}$ to $\mathbb{Z}$?