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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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    $\begingroup$ Elements are not commutative. Elements can commute, and if all elements commute, the operation is commutative. In this case yes, just check all the compositions. $\endgroup$ Aug 26, 2013 at 19:35

4 Answers 4

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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 $1,a,b,ab,ba$.

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  • $\begingroup$ (One can then conclude that all groups of order four are abelian.) $\endgroup$
    – user1729
    Aug 27, 2013 at 18:44
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    $\begingroup$ @user1729 In fact, all groups of order $\le 4$. And since $5$ is a prime, ... $\endgroup$ Aug 27, 2013 at 21:04
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    $\begingroup$ Well, yeah, but you haven't proven that... $\endgroup$
    – user1729
    Aug 27, 2013 at 21:25
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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.

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    $\begingroup$ You can, of course shorten this by observing that the OP's group satisfies the property that $a^2=1$ for all $a\in G$. $\endgroup$ Aug 31, 2013 at 17:57
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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.

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  • $\begingroup$ So, for example would I check that $(12)(34) = (34)(12)$? $\endgroup$
    – user91154
    Aug 26, 2013 at 19:39
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    $\begingroup$ No, that's true and shows you that $(1 \ 2)$ and $(3 \ 4)$ commute, but these aren't elements in your group. You have to show, for example, that $(1 \ 2)(3 \ 4)(1 \ 4)(2 \ 3) = (1 \ 4)(2 \ 3)(1 \ 2)(3 \ 4)$ so that you can conclude that $(1 \ 2)(3 \ 4)$ and $(1 \ 4)(2 \ 3)$ commute. $\endgroup$
    – Jim
    Aug 26, 2013 at 19:41
  • $\begingroup$ oh okay. I see what you mean. $\endgroup$
    – user91154
    Aug 26, 2013 at 19:43
  • $\begingroup$ And, if you check that the product of two distinct nonidentity elements is the third, then there are only two pairs to check. $\endgroup$ Sep 22, 2018 at 12:11
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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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  • $\begingroup$ and $(xy)^2 = e$ because all elements are either identity or of order 2 $\endgroup$
    – Yola
    Nov 18, 2016 at 13:30

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