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Show, without Lagrange theorem, that every group $G$ of order $6$ has an element of order $3$.

I know that if $G$ has an element $a$ of order 6, then $a^2$ has order $3$, but I am not shure how to show that for the case where $G$ does not have an element of order $6$. I also know that we can only have elements $1,2,3,$ or $6$ (we proved this independent without Lagrange theorem).

tmpys
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2 Answers2

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Let $a\in G$. If $a\ne1,$ then $a$ has order $2,3,6$ (as you said). If $a$ has order $6$ then (as you said) $a^2$ has order $3$. If $a$ has order $2$ then we only have two elements in $G$. So let $b\in G$ (not equal to either $a$ or $1$) with order $2$. Note that if $(ab)^2=abab=1,$ then $a(abab)b=a1b\implies ba=ab.$ So we have only $\{1,a,b,ab\}$.

Now add $c$ (suppose it has order $2$ and is not already in $G$). We have at least $5$ elements now: $\{1,a,b,ab,c\}$. We want to show that there are at least $2$ more. Now if it has order $2,$ then $ac$=$ca$ (by the same argument for $ab$). Clearly $ac\ne 1,a,ab,c$. Suppose that $ac=b$. Then (since $a=a^{-1}$) $c=ab$ but $c\ne ab$. Similarly we can show that $bc$ is not equal to any element already in $G$. So $G$ contains at least $1,a,b,c,ab,ac,bc$.

Fred
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  • I don't why a has order 2 implies there are only 3 elements in the group, and if that were true we would be done.....(G has 6 elements) – tmpys Apr 07 '14 at 02:24
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    Sorry. I am supposing that we start with just $1$ in $G$. We need to find $5$ more elements. Start with $a$, then if $a$ has order $2$ , we only have $1,a, a^{-1}$ in $G$. So we need more elements... – Fred Apr 07 '14 at 02:32
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    If $a$ has order $2$, then $a=a^{-1}$... – Nishant Apr 07 '14 at 02:33
  • Thanks Nishant, I have now edited the answer and comment. – Fred Apr 07 '14 at 02:40
  • should that say "(not equal to either a or e)" ? – tmpys Apr 07 '14 at 02:43
  • I'm sorry I am still confused on why, once we add c, we get this other set, and how do we know that we cant have 6? – tmpys Apr 07 '14 at 02:52
  • I have edited the answer – Fred Apr 07 '14 at 03:40
  • I think we also have abc in that last group, and then we have 8 elements, and if we keep doing this we will always get 2^n elements for some n. – tmpys Apr 07 '14 at 07:24
  • Yes but you only need to show that its not possible to have exactly five elements of order two along with the identity. Therefore there must be an element of order three. – Fred Apr 07 '14 at 07:37
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If all nonidentity elements of a group have order $2$, then the group is Abelian. You can then show that any finite abelian group where every nonidenity element has order $2$ must have order a power of $2$.

Nishant
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