If ker $f = \{ (1), (12)(34), (14)(23), (13)(24) \}$, which is the Klein 4 group, how do I prove that it is an Abelian group? Do I just show that each element in the ker $f$ is commutative to prove this?
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$\begingroup$Let $G$ be a non-abelian group. Then $ab\ne ba$ for some $a,b\in G$. Then clearly $a\ne 1$ and $b\ne 1$ and $a\ne b$. Also $ab\notin\{1,a,b\}$ and $ba\notin\{1,a,b\}$. That already makes five distinct elements $1,a,b,ab,ba$.
$\endgroup$ 3 $\begingroup$You don't need to prove it for this particular group, because ALL 4-element groups are abelian, you can prove that like this:
Suppose that for all $a\in G$, we have that $a^2=1$, or equivalently $a=a^{-1}$, then the group is abelian:
$$(ab)=(ab)^{-1}=b^{-1}a^{-1}=ba$$
If our supposition is not true, then there exists an element $a$ such that $a^2\neq 1$ different from $1$, and the group generated by this element, by Lagrange's theorem, must have order $4$, so $\langle a\rangle=G$, so the group is cyclic and abelian.
$\endgroup$ 1 $\begingroup$There are only $4$ elements, so just check by hand that $ab = ba$ for any $a, b$. It's true automatically if either $a$ or $b$ is $1$ so actually there are only three pairs to check.
$\endgroup$ 4 $\begingroup$This is a slightly less direct approach--one you probably would not think of if you were attempting a problem like this for the first time)--but hopefully you (or others) will find this alternative solution to be instructive.
Theorem: Let $G$ be a group such that every element of $G$ other than the identity has order $2$. Then $G$ is abelian.
Proof: Let $x$ and $y$ be any two elements of $G$. Then we must show that $xy=yx$. I will write $e$ to indicate the identity element. $$e=(xy)^2=xyxy$$ $$x^{-1}=x^{-1}e=x^{-1}xyxy=yxy$$ $$y^{-1}x^{-1}=y^{-1}yxy=xy$$ Now note that since $x$ and $y$ each have order $2$, $x=x^{-1}$ and $y=y^{-1}$, so $yx=xy$.
I leave it up to you to verify that the non-identity elements of the group in question have order $2$.
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