4
$\begingroup$

Let $\phi : G_1 \rightarrow G_2$ be a surjective group homomorphism. Let $H_1$ be a normal subgroup of $G_1$ and suppose that $\phi (H_1) = H_2$. Prove or disprove that $G_1/H_1 \cong G_2/H_2$.

I say they are indeed isomorphic. Because:

Let $f$ be the group homomorphism from $G_1$ to $G_2/H_2$ that sends $a$ to $\phi(a)$. Then the kernel of $f$ is everything that is sent to $H_2$. Well by assumption this is $H_1$. Since $\phi$ is surjective, so is $f$, so by the first isomorphism theorem, $G_1/H_1$ is isomorphic to $G_2/H_2$

Is this correct reasoning?

$\endgroup$
  • 2
    $\begingroup$ The kernel of $f$ could be bigger than $H_1$. You only know that $\phi(H_1)=H_2$, but why can't you have $\phi(H_0)=H_2$ with $H_1\subset H_0$? $\endgroup$ – Dan Rust Nov 19 '14 at 2:46
4
$\begingroup$

No your reasoning is incorrect as the comments have already stated.

As a simple counter example take $G_1 = \mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 $, $H_1 = 1 \times 1\times \mathbb{Z}_2$ and then have $G_2 = \mathbb{Z}_2$, with $H_2$ trivial and the map being projection onto the first coordinate. Then clearly $G_1 / H_1 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ and is not isomorphic to $G_1 / H_1$.

$\endgroup$
7
$\begingroup$

Often times before trying to prove something, it is helpful to see if the result is true for a few simple examples. In this instance, try letting $G_2$ and $H_1$ both be trivial to see that this result will not hold in general.

$\endgroup$

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.