3
$\begingroup$

Let $f:G \to H$ be surjective group homomorphism. Show that if $G=\langle a \rangle$, then $H=\langle f(a)\rangle$

Homorphism means $f(a*a)=f(a)*f(a)$

onto means $$\forall h \in H, \exists g \in G : f(g)=h $$

$H=\langle f(a)\rangle \Leftrightarrow H \subset \langle f(a)\rangle, \langle f(a)\rangle \subset H$

Assume $G=\langle a\rangle$

$\Rightarrow$ (if $h \in H$ is $h \in \langle f(a)\rangle$?)

$h \in H$ so $\exists g \in G$ such that $f(g)=h$ and $g\in \langle a \rangle$ so $\exists k \in Z$ s.t $f(a^k)=f(a)^k=h$ so $h\in \langle f(a)\rangle$

$\Leftarrow$ Showing $\langle f(a)\rangle \subset H$)

let $x\in \langle f(a)\rangle $ so $\exists k ...$ $f(a)^k=x$ so $f(a^k)=x$ and $a^k \in G$ that is $f(a^k)\in H$

Critique?

$\endgroup$
  • 3
    $\begingroup$ Use $\langle\rangle$ for $\langle\rangle$. $\endgroup$ – Shaun Sep 28 '17 at 15:05
  • $\begingroup$ Your proof seems fine to me. $\endgroup$ – Shaun Sep 28 '17 at 15:08
  • 1
    $\begingroup$ The second step doesn't require any proof : $\langle f(a)\rangle \subset H$ holds by definition. You might also be interested in this question : math.stackexchange.com/questions/1123005/… $\endgroup$ – Arnaud D. Sep 28 '17 at 15:10
  • $\begingroup$ The only critical remark I can make is that "cruitique" is normally spelled c-r-i-t-i-q-u-e.! $\endgroup$ – Robert Lewis Sep 28 '17 at 15:10
  • 2
    $\begingroup$ Perhaps you could reconsider your username? $\endgroup$ – Alex Clark Sep 28 '17 at 15:21
6
$\begingroup$

The idea of your proof is correct. Good job. :) As said in the comments though, the bit about proving that $\langle f(a) \rangle \subset H$ is unnecessary. I suppose you can write it down if you want, but it's kinda "obviously true": $H$ is the codomain of the map, so $f(a) \in H$, and since $H$ is a group any power of $f(a)$ must be in there too.

In line with what users are saying in the comments, my critiques are about language/grammar. A major thing to remark on is that your proof looks like some math symbols with a few conjunction words thrown in. First and foremost, a proof should be a collection of sentences. Sentences that are grammatically correct, and that give your proof a sense of flow. Remember that your proof is meant to be read and understood by a person, and not just some grading robot looking for correctness ;) . I might write-up the same proof like this:

Assume $G=\langle a \rangle$, and consider some $h \in H$. Since $f$ is surjective, there must be some $g \in G$ such that $f(g)=h$. The group $G$ is generated by $a$, so there must be some integer power $k$ such that $g=a^k$. But $f$ is a homomorphism, so we have $$h = f(g) = f(a^k)=f(a)^k\,,$$ so $h\in \langle f(a)\rangle\,.$ That is to say, a general element $h \in H$ is a power of $f(a)$, so $H = \langle f(a) \rangle$.

Replacing certain mathematical symbols in lieu of their English language counterpart is often a good idea to help the readability of a proof.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.