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May 30th, 2017, 12:46 AM   #1
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True or False

In a Group $G$ if $\displaystyle a \in G$ , $a^7=e$ and $a^9=e$ then $a=e$ ?

I guess it is true. Someone help!

Thank you
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May 30th, 2017, 01:28 AM   #2
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$a = e^5a = (a^7)^5a = a^{36} = (a^9)^4 = e$
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May 30th, 2017, 04:55 AM   #3
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In a group, if $a^n=e$, then the order of a divides n -- if d is the order write $n=dq+r\text{ with }0\leq r<d$. So $e=a^{n-dq}=a^r$. Thus r=0.

So if $a^m=a^n=e$, the order of a divides the gcd of m and n. In particular, if m is prime to n, the order of a is 1; i.e. a=e.
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