# Prove or disprove that $G_1/H_1 \cong G_2/H_2$

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?

• 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 $H_1\subset H_0$? Nov 19, 2014 at 2:46

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.

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 $G_1 / H_1$.

I want to answer this question with a relatively complete discussion. Instead of investigating different situations, we can use the following lemma:

Lemma: Suppose $$\phi$$ is a group homomorphism and $$H \lhd G$$, then $$\frac{G}{\phi^{-1}\circ\phi (H)} \cong \frac{\phi(G)}{\phi(H)}$$

Proof: If we define $$\sigma: G \rightarrow \frac{\phi(G)}{\phi(H)}$$ such that $$\sigma(g)=\phi(g) \phi (H)$$, then $$\sigma$$ is a well-defined onto homomorphism whose kernel is $$\phi^{-1}\circ\phi (H)$$, and the proof is followed by the first isomorphism theorem.

Lemma: Let $$\phi$$ be a group homomorphism. If $$\ker(\phi)\subset H$$, then $$\phi^{-1}\circ\phi (H) =H$$.

Proof: $$H \subset \phi^{-1}\circ\phi (H)$$ is obvius.

Suppose there exists $$g \not \in H$$ such that $$\phi(g) \in \phi(H)$$, then there exists $$h \in H$$ such that $$\phi(g)=\phi(h)$$. This implies $$gh^{-1} \in \ker{\phi}$$, but $$gh^{-1} \not \in H$$, which is a contradiction to the fact that $$\ker(\phi)\subset H$$. Hence, $$\phi^{-1}\circ\phi (H) \subset H$$, and the proof is completed.

Now, if $$\phi$$ is onto then $$\phi(G_1)=G_2$$, and if $$\ker(\phi)\subset H_1$$, then $$\phi^{-1}\circ\phi (H_1) =H_1$$. Hence, acording to $$\phi(H_1)=H_2$$, your result is followed.