# Generator for $\langle a^{21}\rangle \cap \langle a^{10}\rangle$

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.

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 $k=6$.

• 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 $k$. – Rishabh Sareen Jan 14 '17 at 10:40
• @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. – Anurag A Jan 16 '17 at 8:50

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 $a^{|a|}=1$.

• Ohh!! Such a simple solution. Thanks ! – Rishabh Sareen Jan 14 '17 at 10:43

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 $\gcd$.

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