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I tried to prove the following theorem: A group with five elements is abelian.

I know only the definition of a group and a subgroup but no more.(this is a problem from Topics in Algebra by IN Herstein). I tried a few things but to no avail.

Suppose that the group is $G=\{e,g_1,g_2,g_3,g_4\}$. $g_1g_2$ cannot be $g_1$ or $g_2$ or else one of them will be the identity,$e$.So, $g_1g_2$ is either $g_3$ or $g_4$ or $e$.

Case I: $g_1g_2=e$.(so, $g_2g_1=e$). Then $g_1g_3=g_2$ or $g_4$($g_1g_3\not=e$,or else $g_2=g_3$). Say,$g_1g_3=g_2$. That is, $g_1=g_3^{-1}g_2\implies g_1g_2=g_3^{-1}g_2^2=e$ or in other words, $g_2^2=g_3$. Using this, $g_3g_1=g_2^2g_1=g_2(g_2g_1)=g_2$ and we have $g_1g_3=g_1g_2^2=g_2$.

Maybe, this will lead to a solution but I was wondering if there is a much smarter way to do it as this leads to a lot of casework! .

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    $\begingroup$ Easiest way: two facts which you may know already, if not they are quite easy to prove - try it. (1) Any group with a prime number of elements is cyclic. (2) Any cyclic group is abelian. $\endgroup$ – David Aug 26 '14 at 12:52
  • $\begingroup$ What is the order of $g_1$? what is the subgroup generated by $g_1$? it is abelian? $\endgroup$ – Hamou Aug 26 '14 at 12:54
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Proposition 1: Any group of prime order is cyclic.

Let $G$ be a non trivial group of order $p$ and take $g\in G$, $g\neq1$. So $\langle g\rangle$ is a subgroup of $G$, hence its order must divide $p$, which is prime, so $|\langle g\rangle|=p$, hence $\langle g\rangle=G$.

Proposition 2: Any cyclic group is abelian.

A finite cyclic group is by definition of the type $G=\langle g\rangle:=\{1,g,g^2,\dots,g^{n-1}\}$. The fact its elements commute follows easily.

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  • $\begingroup$ My feeling is that this is the best solution. $\endgroup$ – Tomek Kania Aug 26 '14 at 20:25
  • $\begingroup$ Many thanks!!! :-) $\endgroup$ – Joe Aug 26 '14 at 20:26
  • $\begingroup$ A beautiful and simple solution indeed! $\endgroup$ – user96343 Aug 30 '14 at 15:42
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    $\begingroup$ But your proof implicitly uses the Lagrange's theorem (in proof of proposition1) $\endgroup$ – Siddharth Joshi Jul 27 '17 at 14:56
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    $\begingroup$ You haven't read the question carefully. You only have definition of groups and subgroups at your disposal. $\endgroup$ – reflexive Mar 3 '18 at 15:23
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Hint: suppose your group is not abelian. Then you can find two different elements, say (after renumbering) $g_1$ and $g_2$, not equal to the identity, such that $g_1g_2 \neq g_2g_1$. Then the elements $\{e, g_1,g_2, g_1g_2, g_2g_1\}$ are all different. Now try to derive a contradiction (look at $g_1^2$ - which element of the set is this? Do the same for $g_1g_2g_1$).

Below are more details.

In a finite group, there's no element such that $g_1^2=g_1$. Also $g_1^2$ cannot be neither $g_1g_2$ nor $g_2g_1$, because $g_2\neq g_1$. Since $g_2=g_1^2$ would yield $g_1g_2=g_1^3=g_2g_1$, it's imposible, too. Therefore $g_1^2=e$. We now observe $g_1g_2g_1$. Note that $g_1,\, g_1g_2,\, g_2g_1\neq e$, with operating on both left and right, $g_1g_2g_1\neq g_1g_2,\, g_2g_1,\, g_1$. However, if $g_1g_2g_1=e$, we'll have $g_1g_2=g_1^{-1}=g_2g_1$; if $g_1g_2g_1=g_1$, $g_1g_2=e=g_2g_1$ will be produced. A contradiction is hence derived.

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    $\begingroup$ Anyone who knows group theory could have answered this by referring to Lagrange’s theorem. I like this answer for keeping it completely elementary $\endgroup$ – ziggurism Aug 26 '14 at 15:17
  • $\begingroup$ This is beautiful! Very neat trick. $\endgroup$ – Cameron Williams Feb 6 '15 at 5:28
  • $\begingroup$ Does this also use the fact that the set $\{g^n: n \in N\}$ is a subgroup of G, i.e. $g_1^2 \in G$? $\endgroup$ – ensbana Mar 26 '18 at 19:18
  • $\begingroup$ It uses the fact that a group is closed under its algebraic operation, so yes. $\endgroup$ – Nicky Hekster Mar 26 '18 at 20:00
  • $\begingroup$ This is the most elementary proof! I like it because it doesn't have any heavy machinery behind it. $\endgroup$ – kelvin hong 方 Jul 29 at 8:18
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It depends on what facts you have available, but you can see this quickly if know the easily deduced fact that in a group $G$ of order $n$ the order of every element is a factor of $n$. So, every element of a group of order $5$ has order $1$ or $5$, and only the identity has order $1$, by definition. Then, any other element $g$ has order $5$, so every element is a power of $g$, and so the group is abelian, and in fact $G \cong \mathbb{Z}_5$. NB this argument only used that $5$ is prime, so it works as well for any prime order.

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It is known that the order of a conjugacy class from a group element divides the order of the group. With $|G|=5$, this results either in one conjugacy class with $5$ elements, or five conjugacy classes with one element. Since the identity element always has its own conjugacy class, the first case is impossbible. Because all conjugacy classes have only one element, is $G$ commutative.

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    $\begingroup$ The fact that the size of a conjugacy class divides the order of the group is even less elementary than the fact that the order of a subgroup divides the order of the group. $\endgroup$ – Dustan Levenstein Aug 26 '14 at 20:48
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    $\begingroup$ You're right, but this non-elementary fact gives a short proof, which IMHO is favourable in algebra. $\endgroup$ – Thijs Aug 27 '14 at 7:47
  • $\begingroup$ but clearly inappropriate for this particular question. $\endgroup$ – Dustan Levenstein Aug 27 '14 at 13:21
  • $\begingroup$ It is not the simplest proof, but it is an original and different one. Thanks. $\endgroup$ – Hilder Vítor Lima Pereira Sep 25 '17 at 21:32

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