I am new to abstract algebra, group theory. So, its difficult for me to solve this question. Kindly, please help me.
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1what equation is this? – Wonder Jun 26 '14 at 16:43
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Sorry, question edited. – Jun 26 '14 at 16:45
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1Can you show that if $x=a$ is a solution, then so are $x=a^2, a^3,a^4$? – Jyrki Lahtonen Jun 26 '14 at 16:47
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1Jyrki, Yes, I can show this. Then, after that ? – Jun 26 '14 at 16:58
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Let $a^5 = e$. If $a^i = e$ for $1 < i < 5$, then note the fact that $\gcd(i, 5) = 1$. So there exist $p, q$ such that $pi + 5q = 1$; In other words, $a^{pi}\cdot {a^{5q}} = a^1$. As $a^i = a^5 = e$ and any power of $e$ is also $e$, this means that $a = e$. This is excluded since we are looking at non-identity elements as $a$.
In other words, if $a$ is a non-identity element satisfying $a^5 = e$, then $5$ is the lowest exponent for which $a^i = e$.
Use this to get that $a^2, a^3, a^4$ are also non-identity elements which satisfiy this relation.
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