Kernel of a homomorphism is closed under the inverse mapping

Let $G,G'$ be two groups and $f:G \rightarrow G'$ a group homomorphism. How can we show that $H = \ker(f)$ is closed under the inverse mapping?

Actually, I could not understand the statement completely. Any hint would be very helpful.

• Do you mean that to prove $a,b\in \ker f \implies ab \in \ker f$? Commented Nov 27, 2016 at 4:10
• This statement is from Lang's Algebra. Unfortunately, it is something different than the implication you wrote. Commented Nov 27, 2016 at 4:15
• Can you give a page reference for Lang? There does seem to be some ambiguity about what you're asking as noted by @AlanWang.
– Nico
Commented Nov 27, 2016 at 4:17
• It is on 11st page, 3rd edition. Commented Nov 27, 2016 at 4:18
• Right, so you are trying to prove, essentially, that $a\in\ker f\Rightarrow a^{-1}\in\ker f.$ In other words, that $\ker f$ is closed under inverses. Basically, Lang is checking the group axioms to show that $\ker f$ is a group.
– Nico
Commented Nov 27, 2016 at 4:22

1 Answer

As you seem to still be seeking the answer, I will write up an answer for you closely following Lang:

Let $G, G'$ be groups and $f:G\to G'$ be a group homomorphism.

Proposition: $a\in\ker f\Rightarrow a^{-1}\in\ker f$.

Proof: Let $a\in \ker f$ and $e\in G$ and $e'\in G'$ be the identity elements (or units) in their respective groups. By the definition of $\ker f$, we know $f(a)=e'$ and by the definition of a group homomorphism, we know that $f(e)=e'$.

Now compute $$e'=f(e)=f(a\cdot a^{-1})=f(a)f(a^{-1})=e'\cdot f(a^{-1})=f(a^{-1}).$$

Therefore $f(a^{-1})=e'$, so $a^{-1}\in\ker f$, as desired.

From here, you would have to prove the following two statements to be able to conclude $\ker f\le G$:

1. $a,b\in\ker f\Rightarrow ab\in\ker f$
2. $e\in\ker f$.
• Should not we show that $H=ker f$, $H^{-1} \subseteq H$? Commented Nov 27, 2016 at 10:13
• So the first statement, $H=\ker f$, is basically a statement about naming. It's just saying "Let $H$ be the set $\ker f$." Nothing to prove there. The second statement, $H^{-1}\subseteq H$ is what we proved above. :) Notice we took the inverse of something in $H$ and showed it was also in $H$.
– Nico
Commented Nov 27, 2016 at 18:55
• So why $H^{-1} \subseteq H$ is not enough? We are asked to show that it is closed under inverse image, but then we check 2 more: $HH \subseteq H$ and $1 \in H$. Commented Nov 28, 2016 at 18:41
• Well, you only asked for clarification of Lang's proof for closure under inverses. I was just adding that if you wanted to finish the rest of the proof that $H$ is a subgroup of $G$, you would have to prove the other two facts as well. That is what Lang intended the reader to do.
– Nico
Commented Nov 29, 2016 at 4:40