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Suppose that $|a|=24$. Find a generator for $\langle a^{21}\rangle \cap \langle a^{10}\rangle$.

This is one of the question given in Gallian. My instructor gave the following solution $\langle a^{21}\rangle= \langle a^{\gcd(21,24)}\rangle=\langle a^3\rangle$ and similarly $\langle a^{10}\rangle= \langle a^{\gcd(10,24)}\rangle=\langle a^2\rangle$ And later, he went on to prove that the generator of intersection is $a^{6}$.

I didnt quite understand, why he took gcd of $21$ and $24$. I mean what is the result he used. The only result which I can think of is about the generator of cyclic group which states that: $a^k$ generates a cyclic group $G=\langle a\rangle$ of order $n$ iff $\gcd(k,n)=1$

I would be highly thankful if someone can give an insight into this result and can illustrate how this result is applied to the above problem.

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Another (similar) approach

The order of $a^k$ is given by $$|a^k| = \frac{|a|}{\gcd(|a|,k)}.$$ Thus $$|a^{21}| = \frac{24}{\gcd(24,21)}=\frac{24}{3}=8.$$ So $$\langle a^{21} \rangle=\langle a^{3} \rangle$$ Likewise $$|a^{10}| = \frac{24}{\gcd(24,10)}=\frac{24}{2}=12.$$ So $$\langle a^{10} \rangle=\langle a^{2} \rangle$$ You want $$\langle a^k \rangle =\langle a^{21} \rangle \cap \langle a^{10} \rangle = \langle a^{3} \rangle \cap \langle a^{2} \rangle.$$ Thus $k=6$.

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  • $\begingroup$ Thanks Anurag. I was precisely trying to use this result. But couldn't understand as to how can we write $<a^{21}>=<a^{3}>$. So is this by applying fundamental theorem of cyclic groups which says that for each positive divisor $k$ of $n$ (where n is the order of cyclic group G) the group $<a>$ has exactly one subgroup of order $k$. $\endgroup$ – Rishabh Sareen Jan 14 '17 at 10:40
  • $\begingroup$ @RishabhSareen Yes that is one way of looking at it. Another way is to realize that $a^{21}=(a^3)^7 \in \langle a^3 \rangle$, thus $\langle a^{21} \rangle \subseteq \langle a^3 \rangle$ and so on. $\endgroup$ – Anurag A Jan 16 '17 at 8:50
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Maybe this result $\langle a^m\rangle = \langle a^{gcd(m,|a|)}\rangle$. It's not hard to prove.

The inclusion $\langle a^m\rangle \subset \langle a^{gcd(m,|a|)}\rangle$ is immediate, the other one is not so much. Notice $\exists x,y \in \mathbb{Z}$ s.t. $xm+y|a|=gcd(m,|a|)$, then $a^{gcd(m,|a|)}=a^{xm+y|a|}=(a^m)^x\in \langle a^m\rangle$, since $a^{|a|}=1$.

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    $\begingroup$ Ohh!! Such a simple solution. Thanks ! $\endgroup$ – Rishabh Sareen Jan 14 '17 at 10:43
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Recall that $\gcd(m,n) = \min\{mx+ny >0 | x,y\in\Bbb Z\}$. So your cyclic group's generator would be powers of the smallest positive power of $a$ present since it is cyclic. But then this is exactly the definition of the $\gcd$.

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  • $\begingroup$ Ohh.. I failed to realize that. Thanks for such a simple solution. $\endgroup$ – Rishabh Sareen Jan 14 '17 at 10:43

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