0

Heyall, would appreciate some help with abstract algebra because my undergrad brain is fried from doing all the proofs my prof asked me to do. I've hit a bit of a wall with this one; it involves group homomorphisms - super basic but the proof has got to be quite sophisticated cuz my mind is blank.

there are two groups $G$ and $F$ and a mapping $\phi : G \rightarrow F$

the kernel is defined as

$ker (\phi ) := \left \{ x\in G : \phi (x) = e_{H} \right \}$ whereby $e_H$ is the identity element in $H$

to prove: $\phi$ is injective iff $ker(\phi) = \{e_{G}\}$ whereby $e_G$ is the identity element in $G$

would appreciate any tips too, thanks a bunch xx

Aaronv
  • 37
  • 1
    You have to prove two implications. Break it into pieces and review the definitions. These style of problems are what I call a "definition check" because they're checking to make sure you know what all the words mean. Just review that and you should be good to go. – CyclotomicField May 27 '20 at 00:35
  • @AnuragA it doesn't quite hit the nail on the head but that's pretty much what I'm dealin with, thanks – Aaronv May 27 '20 at 00:47
  • 1
    @CyclotomicField that's it, I gotta review the dreaded definitions, thanks for the targeted advice, it brings back my confidence – Aaronv May 27 '20 at 00:49

1 Answers1

2

  • Suppose $\phi$ is injective. As $\phi$ is a group homomorphism, $\phi(e_G) = e_H$, right? So, $g \in G$ is in the kernel of $\phi$ if and only if $\phi(g) = \phi(e_G)$, and by injectivity, $g$ is in the kernel if and only if $g = e_G$.

  • Assume that $\ker \phi$ consists only of $e_G$. To prove the injectivity, suppose that two elements of $G$, $g$ and $h$, satisfies that $\phi(g) = \phi(h)$ and we want to prove that $g = h$. But see this: $$e_H = \phi(g)\phi(h)^{-1} = \phi(gh^{-1})$$ and so, $gh^{-1} \in \ker \phi$, that is, $gh^{-1} = e_G$, and then, $g = h$.

azif00
  • 20,792
  • yoo that was so quick, thank you for taking the time to write it out. it's precisely what I need, I'll just go through it a couple times to commit it to memory. – Aaronv May 27 '20 at 00:51
  • You don't want to commit it to memory. You want to understand it well enough so that you can come up with it yourself from scratch. – Robert Shore May 27 '20 at 01:29
  • yea that's what I mean, but you gotta be creative with these proofs ngl – Aaronv May 27 '20 at 10:36