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This question is from the book Abstract Algebra by Gregory Lee.

What I got so far is:

  • Abelian group is commutative.
  • The answer key says suppose $a$ and $b$ are elements of this group $G$ such that they are distinct (a.k.a. $a\neq b$) and $|a|=|b|=2$

Please help me to show this.

Thanks.

Nick Nguyen
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    See https://math.stackexchange.com/questions/268873/a-finite-group-of-even-order-has-an-odd-number-of-elements-of-order-2 – lhf Dec 11 '20 at 10:44
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    For Abelian groups, the proof is simpler: notice that the set of elements of order $1$ or $2$ is a subgroup (due to $(xy)^2=x^2y^2$), which would end up being a subgroup of order $5$. –  Dec 11 '20 at 10:46
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    2 doesn’t divide 5 – Calvin Khor Dec 11 '20 at 10:49
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    It would be more interesting to prove that no group can have exactly four elements of order $2$. – Derek Holt Dec 11 '20 at 11:00
  • See https://math.stackexchange.com/questions/3796395/why-a-group-that-has-four-elements-of-order-two-does-not-exist and https://math.stackexchange.com/questions/740397/the-number-of-elements-of-order-2-in-an-infinite-group – lhf Dec 11 '20 at 11:05

1 Answers1

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Here is an elementary argument.

If $a$ and $b$ are two elements of order $2$, then there are at least three elements of order $2$: $a,b,ab$.

If $c$ is a fourth element of order $2$, then there are at least seven elements of order $2$: $a,b,c,ab,ac,bc,abc$.

Jacob Antony
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lhf
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