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Show that a nonabelian group must have at least five distinct elements.

I just learn abstract algebra by self study. I want help to solve this problem. Just give me a hint.

Martin Sleziak
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Lingnoi401
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4 Answers4

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You need an instance of $ab\ne ba$. That requires $a\ne b$. Also $a\ne 1$ and $b\ne 1$ as $1$ commutes. Also, $a,b$ are not inverse of each other as those commute. Hence $1, a, b, ab, ba$ are pairwise distinct

Hagen von Eitzen
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    We know the result can be improved to six pairwise distinct elements (because we know all groups of order 5 so well). Is there a "nice" way to see that if we throw in all kinds of other products, like $a^2$, $aba$, and many more, then we can find at least one more of all those which is distinct to each of the five on your list, bringing the group order up to at least 6? – Jeppe Stig Nielsen Oct 16 '16 at 22:13
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    @JeppeStigNielsen Nope. For example, we might have $a^2=1$. There is no single specific expression that we can claim to be different - just check with different choices for $a,b\in S_3$. To exclude order five, I suppose it is ok to use the fact that 5 is a prime. – Hagen von Eitzen Oct 17 '16 at 07:52
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    @JeppeStigNielsen Not easily. If $a^2\ne 1$, then it is also not $a,b,ab,ba$ either since these all imply $ab=ba$. Similarly for $b^2\ne 1$. If $a^2=b^2=1$, then $aba=1\to ba=a\to b=1$, $aba=a\to ba=1$, $aba=b\to ab=ba$, $aba=ab\to a=1$, $aba=ba\to a=1$. Thus $aba$ is distinct from $1,a,b,ab,ba$. – Mario Carneiro Oct 17 '16 at 07:53
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    @MarioCarneiro So your comment shows that either $a^2$, or $b^2$, or $aba$, will be "new" (not in Hagen's list of five distinct elements), so there are (by choosing the right one of these extra candidates) at least six distinct elements. And $S_3$ is good to consider. – Jeppe Stig Nielsen Oct 17 '16 at 09:32
  • @Jeppe Actually the list can be improved to just $a^2$ or $aba$, because the last part does not use that $b^2=1$ anywhere. – Mario Carneiro Oct 17 '16 at 17:45
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In fact it must have at least $6$ elements.

You can discard the possibilities of the group having exactly $1$ element immediately. You can discard the possibility of the group having a prime number of elements because any such groups are cyclic, so $2,3$ and $5$ are discarded.

It remains to show that no non-abelian group with $4$ elements exists.

If it has an element of order $4$ then it is cyclic, otherwise every element must have order $2$ or $1$. And a group in which this happens is abelian, since $(ab)^2=e=a^2b^2$

Martin Sleziak
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Asinomás
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Alternative solution: Suppose it is not abelian, then it has two elements $a$ and $b$ that do not commute, hence the group contains $e,a,b,ab,ba$ and must have at least $5$ elements.

Asinomás
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Hint: Try to make a list of all groups of orders $1,2,3,4$ (up to isomorphism). There are not many (five to be precise) and you will see, they are all abelian. (One might add that there is also only one of order $5$ and it is also abelian.)

J.R.
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