A group with five elements is Abelian

Question

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! .

Answer 1

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 non-trivial 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.

Answer 2

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.

Answer 3

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.

Answer 4

You're on the right track. There is a "smarter" way to do it, but it requires more knowledge of theorems. Part of the point of this exercise is to help you appreciate those theorems more when you learn them (since you asked the question 6 years ago, I'm assuming you've learned it by now, but I'm answering in case anyone else has the same question).

My solution had three parts: First, assume $a,b\in G$ with $a\neq b$. The first part is to show that $e,a,b,ab,ba$ are all distinct (that's ${5\choose 2} = 10$ equalities you have to disprove). This lets you conclude that these five elements are exactly the five elements of $G$. Next, show that $a$ and $b$ are each their own inverses, and $(ba)^{-1}=ab$. Finally, analyze the element $aba$, and show that you get a contradiction no matter which of the five elements of $G$ is equal to $aba$.

Since it's apparently a fairly common undergraduate problem, I won't give the entire solution, but that outline should give you clear enough guidance that you can fill in the details yourself.