I have a question regarding the application of the First Isomorphism Theorem for Groups in proofs; why are the proofs not concerned with whether the respective map is injective?

To clarify my question, let me state the First Isomorphism Theorem:
Let $G$ and $H$ be two groups and let $\phi: G \mapsto H$ be a homomorphism. Then, the First Isomorphism Theorem states that $ G/ Ker(\phi) \cong Im(\phi) $.

Then, to prove that $ Im(\phi) = H $ we have to show that $ \phi $ is surjective. This implies that $ G/ Ker(\phi) \cong H $.

My question then is: why are we not worried about whether $\phi$ is injective? Isn't an isomorphism a bijective homomorphism? Or is it that the quotient group $ G/ Ker(\phi) $ eliminates the need for the injectivity of $ \phi $?.

I wonder whether the fact that $\phi$ is injective $\implies G/Ker(\phi) = G$ has something to do with my question. Does the First Isomorphism Theorem allow us to say more things about homomorphisms that are not injective i.e whose kernel is not trivial?

Thus, I believe that the First Ismorphism Theorem essentially has three iterations:

  1. $\phi: G \mapsto H$ is a homomorphism $ \implies G/ Ker(\phi) \cong Im(\phi) $.
  2. $ \phi$ is a surjective homorphism $\implies G/ Ker(\phi) \cong H $.
  3. $\phi$ is an injective and surjective homomorphism i.e an isomorphism $\implies G/ Ker(\phi) = G \cong H $

Does my reasoning make sense? Thank you!

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    $\begingroup$ If it were injective you wouldn't need the theorem! $\endgroup$
    – Randall
    Commented Dec 9, 2023 at 15:05
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    $\begingroup$ The First Isomorphism Theorem basically states: "If you take an arbitrary homeomorphism, and you make it surjective (by only considering the image) and you make it injective (by dividing by the kernel), you get an isomorphism". The kernel part is your injectivity, in the same way that the image part is your surjectivity. $\endgroup$ Commented Dec 10, 2023 at 4:53

2 Answers 2


We are worried about injectivity of $\phi$! That's why we take the quotient by the kernel! $\phi$ need not inject on $G$, but the induced map from the quotient $G/\ker\phi$ - where we have killed off all the elements that contradict injectivity of $\phi$ ! - is always injective. This is true if we took $G/N$ for some normal $N$ containing $\ker\phi$ as well; however, if we took a strictly larger quotient then the image of $G/N\to H$ would be strictly smaller than $\mathrm{im}(\phi)$. So, killing off $\ker\phi$ in the quotient is exactly the correct thing to do.

Points in the image of $\phi$ are identifiable with points in $G$... up to a factor of $\ker\phi$, which is worrying (and we are worried) if $\phi$ isn't injective. "The" solution is to set this factor to be zero by passing to $G/\ker\phi$, which now gives a genuine $1$-$1$ identification.


The kernel is the same as the pre-image of the identity element of $H$. For any injective function, the pre-image of all elements has at most one element. So for functions in general, we have to check every element in the range and check whether their pre-image has only one element. For a homomorphism, though, we only have to check one element. That is, if $\phi$ is a homomorphism $G \rightarrow H$, then we can pick any $h$ in $H$ and $\phi$ is injective iff the pre-image of $h$ has at most one element. So if $\phi$ were injective, the kernel would be trivial, and the theorem would reduce to merely saying "A trivial quotient of a group is isomorphic to the original group."

The theorem is interesting only when the kernel is non-trivial, in which case we know that $\phi$ isn't injective. Rather than taking a single element of $G$ to a unique element of $H$, $\phi$ takes equivalence classes (i.e. cosets) to unique elements of $H$. If the kernel is $K$, then the we can write the pre-image of $0_H$ as $0_GK$. What the theorem says is that this is true of any element of $G$: the pre-image of $f(g)$ is the coset $gK$. Basically, $f$ "ignores" any element of $K$. $f(gk) = f(g)$ for any $k \in K$.

The statement $ G/ Ker(\phi) \cong H $ is saying that there is an isomorphism from the quotient space to $H$, but it's not saying that $\phi$ is that isomorphism. Let's call the isomorphism $\psi$. Then the theorem tells us we can get $\psi$ by taking $\psi(gK) = \phi(g)$. Note that while the domain of $\phi$ is $G$, the domain of $\psi$ is the set of cosets. And note that $\psi$ being well-defined isn't completely trivial. It is the isomorphism theorem that says it is in fact well-defined. For $\psi$ to be well-defined, we need that $\phi(g)$ gives the same result for all members of a coset. That is, if $g_2 \in g_1K$, then $\phi(g_2)$ has to be the same as $\phi(g_1)$.

In other words, the theorem is saying that $\phi$ is not injective over the elements of $G$ (after all, elements of the same coset have to go to the same destination), but $\psi$ is injective over the cosets. Each coset is sent to a unique element of $H$. No two cosets get sent to the same element of $H$.


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