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Let $G$ be a group. Let $a$ be an element. Let $n,k$ be pozitive integers. Let $m$ be least common multiple of $n$ and $k$. Prove $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{m} \rangle$. Trying to show both are subsets of each other. One side is trivial. But I can not show that $\langle a^n \rangle$ intersection $\langle a^k \rangle$ is a subset of $\langle a^m \rangle$

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Let $g$ be in the intersection, i.e. $g=(a^n)^r=a^{nr}$ and $g=(a^k)^s=a^{ks}$. This means $a^{nr}=a^{ks}$. Can you take over from here?

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  • $\begingroup$ No this is what I have done already $\endgroup$ – mathematiccian Oct 24 '14 at 9:59
  • $\begingroup$ If nr=ks it is easy but the case they are not equal I can not do $\endgroup$ – mathematiccian Oct 24 '14 at 10:12
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You took $lcm (n,k)=m$

Now, let $x=a^t \in \langle a^n \rangle \bigcap \langle a^k \rangle$

Then, $n$ divides $t$ and $k$ divides $t$.

So, $m=lcm(n,k)$ divides $t$

i.e. $x=a^t\in \langle a^m \rangle$

For opposite containment

Let $x=a^t\in \langle a^m \rangle$

Then, $m$ divides $t$

Now, since $n$ and $k$ divide $m$ you have the containment.

So equality holds.

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  • $\begingroup$ But n does not have to divide t? Take a=1. Take x=1^5 it is an element of 1^7 and 1^8 but 5 does not divide 7 nor 8 $\endgroup$ – mathematiccian Oct 24 '14 at 10:57
  • $\begingroup$ But $1^8=1^7=1^{10}$ and you know, $5$ divides $10$ $\endgroup$ – Swapnil Tripathi Oct 24 '14 at 10:59
  • $\begingroup$ If $a^t\in \langle a^n \rangle$. Then $a^t=(a^n)^k=a^{nk}$. So, $t=nk$ and $n$ divides $t$ $\endgroup$ – Swapnil Tripathi Oct 24 '14 at 11:02
  • $\begingroup$ We can always find a t such that n divides t, And we can always find a t such that k divides t. But both t's does not have to be equal. We need to show t's are equal? $\endgroup$ – mathematiccian Oct 24 '14 at 11:10
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    $\begingroup$ If $a$ has order $12$ then $a^2=a^{50}\in \langle a^{10} \rangle$ but $10$ does not divide $2$. $\endgroup$ – lhf Oct 24 '14 at 11:38
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Consider the canonical surjective homomorphism $\pi:\mathbb Z \to A=\langle a \rangle$ given by $t \mapsto a^t$.

Let $A_t = \langle a^t \rangle$ for $t \in \mathbb N$. All subgroups of $A$ are of this form. The lattice of subgroups of $A$ is isomorphic to the lattice of subgroups of $\mathbb Z$ containing $\ker \pi$.

If the order of $a$ is infinite, then $\pi$ is an isomorphism taking $t\mathbb Z$ to $A_t$. The $A_t$ are all different and the result follows from the fact that the lattice of subgroups of $\mathbb Z$ reflects divisibility. In particular, $n \mathbb Z \cap k \mathbb Z = lcm(n,k) \mathbb Z$.

If the order of $a$ is finite, the subgroups $A_t$ cannot be all different, since there are only a finite number of them. They are all different only when we take $t$ to be a divisor of $N$, the order of $a$. For such $t$, the lattice of subgroups of $A$ is isomorphic to the lattice of subgroups of $\mathbb Z$ that contain $N\mathbb Z$, which reflects divisibility among the divisors of $N$.

In general, $A_t = A_{gcd(N,t)}$. This is the key point. It follows from writing $gcd(N,t)=uN+vt$.

The result then follows from the distributivity property of gcd and lcm: $$ gcd(u, lcm(v, w)) = lcm(gcd(u, v), gcd(u, w)) $$ Indeed, $$ A_n \cap A_k = A_{gcd(N,n)} \cap A_{gcd(N,k)} = A_{lcm(gcd(N,n),gcd(N,k))} = A_{gcd(N,lcm(n,k))} = A_{lcm(n,k)} $$

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