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

  • $\begingroup$ 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. $\endgroup$ – 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 $\Bbb Z$?
  • $\begingroup$ So if it is addition, then the identity element is zero correct? so it would have to be mapped to zero in Z, no? $\endgroup$ – PersonaA Dec 8 '15 at 21:35
  • $\begingroup$ 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? $\endgroup$ – fleablood Dec 8 '15 at 21:43
  • $\begingroup$ Wouldn't it be all of them relatively ptime? $\endgroup$ – PersonaA Dec 8 '15 at 21:47
  • $\begingroup$ But what elements a in Z (not Z/n) have the property an = 0? In Z. Not Z/n $\endgroup$ – fleablood Dec 8 '15 at 21:48
  • $\begingroup$ Only a=0 then ? $\endgroup$ – PersonaA Dec 8 '15 at 21:48

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