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Suppose $\langle a\rangle$ is a cyclic group of order $n$ and $\langle a^m\rangle$ is a (normal) subgroup of $\langle a\rangle$. $a^k$ be any element of $\langle a\rangle$. Now by division algorithm, $k=mq+r$ where $r=0,1,2,...,m-1$. Consider the left coset $a^k\langle a^m\rangle$.

$a^k=a^{mq}a^r$ and since $a^{mq}\in\langle a^m\rangle$, $a^k<a^m>=a^r<a^m>$. Thus there are $m$ distinct left cosets of $\langle a^m\rangle$ in $\langle a\rangle$.

But, $|\langle a\rangle/\langle a^m\rangle|=|a|/|a^m|=\gcd(m,n)$ and this means that $\gcd(m,n)=m$ i.e. $m|n$. But this isn't necessarily true. Where was I wrong?

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  • $\begingroup$ Forgive me for asking such a question. I am new to this concept. $\endgroup$ – Hrit Roy Jan 1 '18 at 14:04
  • $\begingroup$ There could be $1 \leq s \leq m-1$ such that $a^s \in (a^m)$. Consider $n=12, m=8$. Then $a^4 \in (a^8)$, so the coset $a^4(a^8)$ is same as $(a^8)$. $\endgroup$ – Krish Jan 1 '18 at 14:36
  • $\begingroup$ @Krish exactly. I even added the answer a minute ago. I falsely assumed that there could be m cosets $\endgroup$ – Hrit Roy Jan 1 '18 at 14:37
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When you write $a^m$ it's not necessarily true that $0\le m<n$. What you can safely say is that the order of $a^m$ (which is the same as the order of $\langle a^m\rangle$) divides $n$.

What's the order of $a^m$? You have to find the minimal positive integer $l$ such that $(a^m)^l=1$ (or $e$, if you're used to name this way the identity of the group). Now $a^{ml}=1$ if and only if $ml$ is a multiple of $n$. Hence, the minimum such $l$ is so that $ml$ is the least common multiple of $m$ and $n$. So we need $(ml)d=mn$, where $d=\gcd(m,n)$, which is the same as $l=n/d$.

In a compact formula, the order of $a^m$ is $n/\!\gcd(m,n)$.

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$\langle a^m\rangle$ will be a group if the identity $e\in\langle a^m\rangle$. This will be true if and only if $m\mid n$.

Use $a^n=e$ and $n$ is the smallest integer satisfying this property.

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  • $\begingroup$ If $|a|=3$ then $|a^2|=3$. So $<a>=<a^2>$ but 2 does not divide 3. $\endgroup$ – Hrit Roy Jan 1 '18 at 14:26
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I figured it out. It was a silly mistake. When I said that there were $m$ cosets, I assumed right then that $m|n$ because only then would the elements $a,a^2,...,a^{m-1}$ not be in the subgroup.

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