# Few group theory questions

I am trying to solve the following;

First, given $$G$$ is a group and $$H$$ a subgroup of $$G$$, what can we say about the relation $$a \cong b$$ if $$b^{-1}a \in H$$

I can show that it is reflexive as the identity is always in the subgroup.

if $$a \cong b$$ then $$b^{-1}a \in H$$ and so $$(b^{-1}a)^{-1}=a^{-1}b \in H$$ so $$b \cong c$$.

Now I must determine if $$G$$ being abelian is required for this to be transitive.

My thought would be that if $$a \cong b$$ then $$b^{-1}a \in H$$ and if $$b \cong c$$ then $$c^{-1}b \in H$$ and by properties of subgroups so to is $$c^{-1}bb^{-1}a=c^{-1}a$$ which would seem to imply $$a \cong c$$, but I don't know if I am somehow doing something that couldn't be done if $$G$$ was not abelian. What do you guys think?

And second part asks to discuss the possible homomorphism (group) from $$\mathbb{Z}/n\mathbb{Z}$$ to $$\mathbb{Z}$$ and it says for $$n \ge 1$$, is every homomorphism the zero one. Either yes, no, or it depends on $$n$$.

My thoughts are that only the zero homomorphism is possible, since if $$\phi$$ was a homomorphism then $$\phi(ab)=\phi(a)\phi(b)$$ and this would quickly result in having $$ab=n=0$$ but $$\phi(a)$$ and $$\phi(b)$$ not being $$0$$. If $$n$$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?

Thank you all

• I see nothing wrong with your transitivity proof and it doesn't require commutativity. In part 2 you seem to be confabulating addition with multiplication. This is group, not field theory. ab = n = e means a and b are inverses. Not that one or the other is 0. – fleablood Dec 8 '15 at 21:32

Let $$\phi:\Bbb Z/n\to \Bbb Z$$ be a homomorphism.

So $$\phi(a + b) = \phi(a) + \phi(b)$$ so $$\phi(a) = \phi(0 + a) = \phi(0) + \phi(a)$$ so $$\phi(0) = 0$$.

Now \begin{align}\phi(am) &= \phi\left(\underbrace{a + a + \dots + a}_{m\text{ times.}}\right) \\ &= \underbrace{\phi(a) + \phi(a) + \dots + \phi(a)}_{m\text{ times.}} \\ &= \phi(a)\times m.\end{align}

And $$an = 0$$ for all $$a$$ in $$\Bbb Z/n$$.

So $$\phi(an) = \phi(a)\times n = 0$$ for all $$a$$ in $$\Bbb Z/n$$. But in $$\phi(a) \in \Bbb Z$$. $$\phi(a)\times n = 0 \implies \phi(a) = 0$$. This is true for all $$a$$ in $$\Bbb Z/n$$.

So $$\phi$$ is the zero homomorphism. It's the only possible homomorphism.

1. Correct, commutativity is not needed and not used in your proof.
2. The group operation is rather addition.
Hint: Where can the equivalence class $[1]_{n\Bbb Z}$ be mapped by a homomorphism to $\Bbb Z$?
• So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no? – PersonaA Dec 8 '15 at 21:35
• The identity gets mapped to the identity under all homomorphisms to any group. Were can 1, the generator of Z/n, get mapped to, is what is being asked. Hint: 1n = 0 in Z/n. So which a $\in$ Z have the property an = 0? – fleablood Dec 8 '15 at 21:43
• Wouldn't it be all of them relatively ptime? – PersonaA Dec 8 '15 at 21:47
• But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n – fleablood Dec 8 '15 at 21:48
• Only a=0 then ? – PersonaA Dec 8 '15 at 21:48