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?

  • 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, 2014 at 2:46

3 Answers 3


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.


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 $G_1 / H_1$.


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.


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