# Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective

I am solving the following exercise:

Let $\phi : G_1 \rightarrow G_2$ be a homomorphism (where $G_1$ and $G_2$ are groups) and $\ker \phi := \{ g \in G_1 \mid \phi(g) = e \}$

now I have to prove that

a) $\ker \phi$ is a subgroup of $G_1$,

b) $\phi$ is injective if and only $\ker \phi = \{ e \}$

My Problem: Until yet we have not really covered the topic of homomorphism between groups in our lectures. Anyhow I looked it up on wikipedia and found the definition for a subgroup as the following: $(U, \circ)$ is a subgroup of $(G, \circ)$ if $U$ is not an empty set. Therefore:

• $a,b \in U \Rightarrow a \circ b \in U$
• $a \in U \Rightarrow a^{-1} \in U$
• $a,b \in U \Rightarrow a \circ b^{-1} \in U$

so I began to work with these definitions. I somehow managed to prove what I'm supposed to but I'm not sure if I did it the right way. I would be very thankful about some additional words to my attempt and also corrections. Thank you a lot in advance.

My Attempt:

a) $\ker \phi$ is a subgroup of $G_1$. So we can take two elements $x,y \in \ker \phi$ which are $x := \phi(g_1) = e$ and $y := \phi(g_2) = e$ and show that $x^{-1}$ and $x \circ y$ $\in$ $\ker \phi$.

Since $x\circ x^{-1} \in \ker \phi$ we can say: $x\circ x^{-1} = e \ \Leftrightarrow \ \overbrace{\phi(g_1)}^{= \ e} \circ x^{-1} = e \ \Rightarrow \ x^{-1} = e \ \Rightarrow \ x^{-1} \in \ker \phi$.

It must also be true that $x \circ y \in \ker \phi$ this is easily shown by: $x \circ y \ \Leftrightarrow \ \overbrace{\phi(g_1)}^{= \ e} \circ \overbrace{\phi(g_2)}^{= \ e} = e \ \Rightarrow \ x \circ y \in \ker\phi$

b) To show that $\phi$ is injective when $\ker\phi = \{ e_{G_1} \}$ we must show that $\ker\phi = \{ e_{G_1} \}$ has only one fiber which then has to be $\phi^{-1}(e_{G_2})$.

So we can take two elements $g_1,g_2 \in G_1$ and if $\phi(g_1) = \phi(g_2) = e \ \Rightarrow \ g_1 = g_2$ we can state that $\phi$ with $\ker \phi = \{e\}$ is injective.

• I think your proof that $x^{-1}\in\ker\phi$ is wrong. If $x^{-1}\in\ker\phi\subseteq G_1$ and $\phi(g_1)\in G_2$, then I don't think you can necessarily apply a common group operation. Also, you seem to miss the "only if" for part b. – Robin Goodfellow Oct 14 '14 at 10:31
• Hey Robin could you be a little more precise or in detail because I don't understand what you mean. I mean since $e$ is the only element in our kernel the inverse $e$ again must also be in it? Or do I misunderstand something – Mainviel Oct 14 '14 at 10:33
• There is definitely some confusion about which group things live in - for example, in your part a), you take $x\in\ker{\phi}$ such that $x=\phi(g_1)$ - but $\ker{\phi}\subseteq G_1$, and $\phi(g_1)\in G_2$ for any $g_1\in G_1$, so this is impossible. What you want is to notice that because $x\in\ker{\phi}$, then $\phi(x)=e_{G_2}$. For similar reasons, you can't write $\phi(g_1)\circ x^{-1}$, as the two elements live in different groups. – mdp Oct 14 '14 at 10:53
• alright I understand my mistake, so if I would write $\phi^{-1}(g_1)$ instead of $\phi(g_1)$ my formulation would be correct? – Mainviel Oct 14 '14 at 10:58
• @Mainviel Don't do that! If $\phi$ isn't injective, which might be the case at that stage, then $\phi^{-1}(g_1)$ is a set with more than one element, which makes things even worse! At the moment you haven't written the correct definition of $x\in\ker{\phi}$; if you fix this, the problem of elements living in the wrong group should go away. – mdp Oct 14 '14 at 11:01

## 1 Answer

Some notes, which hopefully should help you fix things.

1) If $x,y\in\ker{\phi}$, then $x,y\in G_1$, so $x$ and $y$ cannot be in the image of $\phi$, which is a subset of $G_2$. What you know is that $\phi(x)=\phi(y)=e_{G_2}$; so what is $\phi(x\circ y)$? Don't forget to show that the kernel is non-empty - it contains $e_{G_1}$.

2) $\phi(g_1)\circ x$ is undefined, because $\phi(g_1)$ is an element of $G_2$, whereas $x$ is an element of $G_1$ - always be careful not to do this.

3) For the second statement ("$\ker{\phi}=\{e_{G_1}\}$ if and only if $\phi$ is injective") it might help you to do each direction separately. Firstly, if $\phi$ is injective, and $\phi(g)=e_{G_2}$ then what must we be able to say about $g$? (Hint: what is $\phi(e_{G_1})$?). Secondly, assume $\ker{\phi}=\{e_{G_1}\}$; if $\phi(g_1)=\phi(g_2)$, what is $\phi(g_1g_2^{-1})$?

Please comment if anything is unclear!

• Yes, that's much better! You should also check that $\phi(x^{-1})=e_{G_2}$ if $\phi(x)=e_{G_2}$ (and that $\ker{\phi}$ isn't empty, but this should be easier!). – mdp Oct 14 '14 at 11:30
• For 1) $\phi(x \circ y) = e_{G_2}$ since $\phi$ is an homomorphism we can say: $\phi(x \circ y) \Leftrightarrow \overbrace{\phi(x)}^{= \ e_{G_2}} \circ \overbrace{\phi(y)}^{= \ e_{G_2}} = e_{G_2}$ For 3) 1st: $\phi$ injective $\phi(g) = e_{G_2}$ now we can say that $g = e_{G_1}$ since $\phi(e_{G_1}) = e_{G_2} = \phi(g)$ 2nd: Since $\phi$ is a homomorphism and $\phi(g_1) = \phi(g_2)$ so we know $g_1^{-1} = g_2^{-1} \Rightarrow \phi(g_1 \circ g_1^{-1} = \phi(e_{G_1}) = e_{G_2}$ is that somehow true? – Mainviel Oct 14 '14 at 11:32
• So for $x^{-1} \in \ker\phi$ we can say $x \circ x^{-1} = e_{G_1}$ now we can apply $\phi$ so we get: $\phi(x \circ x{-1}) = \overbrace{\phi(e_{G_1})}^{= \ e_{G_2}} \Leftrightarrow \overbrace{\phi(x)}^{= \ e_{G_2}} \circ \phi(x^{-1}) = e_{G_2} \Rightarrow \phi(x^{-1})= e_{G_2}$. If the kernel was empty we would not have an injective function and therefore not an homomorphism. Or how do I prove this? Thank you a lot for your help! – Mainviel Oct 14 '14 at 11:52
• The proof for the inverse is correct. Homomorphisms don't have to be injective, so that wasn't the right thing to say - the kernel is non-empty because it contains $e_{G_1}$; you even write this down in the proof that $\phi(x^{-1})=e_{G_2}$! – mdp Oct 14 '14 at 12:07
• well then the proof is 'only' saying: that $e_{G_1}$ is in the kernel ? That somehow don't look like a proof to me. I mean the fact is given in the task setting. Thank you alot! – Mainviel Oct 14 '14 at 12:16