# Homomorphisms and kernels

• Nov 16th 2010, 04:20 PM
kathrynmath
Homomorphisms and kernels
Let c:G1--->G2 and d:G2--->G3 be group homomorphisms. Prove that dc:G1--->G3 is a homomorphism. Prove that ker(c) is a subset of ker(d).

Since c and d are homomorphisms, we know c(ab)=c(a)c(b) and d(ab)=d(a)d(b).
We want to show dc(ab)=dc(a)dc(b)
I get stuck at this point.
• Nov 16th 2010, 05:25 PM
Drexel28
Quote:

Originally Posted by kathrynmath
Let c:G1--->G2 and d:G2--->G3 be group homomorphisms. Prove that dc:G1--->G3 is a homomorphism. Prove that ker(c) is a subset of ker(d).

Since c and d are homomorphisms, we know c(ab)=c(a)c(b) and d(ab)=d(a)d(b).
We want to show dc(ab)=dc(a)dc(b)
I get stuck at this point.

I think you're making this harder than it is. If $a,b\in G_1$ then $c(ab)=c(a)c(b)\in G_2$ and so $d(c(ab))=d(c(a)c(b))=d(c(a))d(c(b))\in G_3$.

I'll let you think about the second part.
• Nov 16th 2010, 05:38 PM
kathrynmath
ker c is defined as x in G1 such that c(x)=e
ker d is defined as x in G2 such that d(x)=e
• Nov 16th 2010, 05:40 PM
Drexel28
Quote:

Originally Posted by kathrynmath
ker c is defined as x in G1 such that c(x)=e
ker d is defined as x in G2 such that d(x)=e

Ok, so what does that tell us?
• Nov 16th 2010, 05:47 PM
kathrynmath
since c(x) is in G2 then ker c is a subset of ker d
I'm not sure
• Nov 16th 2010, 06:41 PM
topspin1617
Well, first, trying to say $\mathrm{ker}c\subseteq \mathrm{ker}d$ doesn't make any sense; the two are not even contained in the same group.

I believe the question should be to show that $\mathrm{ker}c\subseteq \mathrm{ker}(dc)$

Now, just think about it; what would it mean to say $\mathrm{ker}c\subseteq \mathrm{ker}(dc)$?

Purely by definition of what it means to be a subset, it means that we want to show that, if $x\in \mathrm{ker}c$, then $x\in \mathrm{ker}(dc)$ as well. Why should this be true?
• Nov 16th 2010, 07:14 PM
kathrynmath
You are right. Guess I misread the question.
ker dc is defined as x in G1 such that dc(x)=e
Since they both involve G1 it must be a subset?
• Nov 17th 2010, 04:40 PM
kathrynmath
Is that anywhere near the right track? I guess I'm having trouble applying the definition.