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.
Best 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!