# How can I prove that the kernel must include the neutral element?

Given are two groups $(G,+)$, $(G',*)$ with neutral elements $e, e'$, a homomorphism $f: G \to G'$ and a subgroup $H' \leq G'$.

I would like to show that this implies that $H := f^{-1}(H')$ has to be a subgroup of $G$.

This can be done by proving that

1. $H$ contains the (unique) neutral element $e$
2. $H$ contains all inverses of its elements
3. Element concatenations remain in $H$

I'm stuck at (1). My thoughts:

This is about the kernel of $f$. The kernel will include $e$ and $f(e) = e'$. However, we don't know any of that yet because we're not sure if $H$ is a subgroup yet.

I attempted $$\exists x \in H: f(x) = e' = e' * e'^{-1} = f(x) * f(x)^{-1} = f(x + x^{-1}) = f(e) \text{ thus } x=e$$ However, this has to be wrong because it would imply that every f is injective, which does not have to be the case. I am not sure where the mistake is.

I also made a couple of other attempts, none yielded a useful result though.

• @MarkBennet Yes, thanks!
– mafu
Commented Apr 12, 2014 at 15:16
• You are trying to show that $e$ belongs to $H$. To do that, you just need to verify that $f(e)$ belongs to $H'$. Commented Apr 12, 2014 at 15:18
• In general, when you are trying to show some $x$ belongs to $f^{-1}(Y)$, you just need to show that $f(x)$ belongs to $Y$. Commented Apr 12, 2014 at 15:19
• @Braindead $f(e) \in H'$ is given because $H'$ is a subgroup - but why does this suffice?
– mafu
Commented Apr 12, 2014 at 15:19

By definition of group homorphism, $f(e)=e'$. This means precisely that $e\in H$, since, by definition of preimage, $$f^{-1}(H')=\{x\in G:\ f(x)\in H'\}$$ and $e'\in H'$ (as $H'$ is a subgroup of $G'$).

• Sorry, I still do not understand this. Could there not be a $x \in H, x \neq e$ with $f(x)=e'$?
– mafu
Commented Apr 12, 2014 at 15:21
• Of course, but how does this affect the proof that $H$ is a subgroup of $G$? To prove $H\leq G$, you just need to show (among the other things) that $e\in H$, while it is not required that $e$ is the unique element in $H$ mapped to $e'$ via $f$! Commented Apr 12, 2014 at 15:25
• The inverse image contains all such $x$, including the identity. Commented Apr 12, 2014 at 15:26
• My rationale: If there is no $e \in H$, but $\exists x \in H: f(x)=e' \land x\neq e$, then $H$ is not a group yet $H'$ is a subgroup. This would also mean that while $f^{-1}(H')$ contains this $x$, it would still not contain $e$.
– mafu
Commented Apr 12, 2014 at 15:32
• Edit: I see it now! As you wrote, the inverse image considers all elements of $G$, not of $H$, and $e \in G$.
– mafu
Commented Apr 12, 2014 at 15:36