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Let $a$ and $b$ be in a group. If $|a|=10$ and $|b|=21$, show $\langle a \rangle \cap \langle b \rangle = \{e\}$

My rough attempt is as follows.

Notice that $|a|=10$ and $|b|=21$ are relatively prime. Let $a,b\in G$ of order $n$. Let $x \in \langle a \rangle$ . $|a|=10$, then we obtain $x^{10}=e$. Recall $a^k=e \rightarrow |a|$ divides $k$. Similarly, Let $y \in \langle b \rangle$ . $|b|=21$, then we obtain $y^{21}=e$. Now let $A$ be the set of all factors of 10; let $B$ be the set of all factors of 21. Let $g \in \langle a \rangle \cap \langle b \rangle$. Since 10 and 21 are relatively prime, then $\langle a \rangle \cap \langle b \rangle =\{1\}$. It follows that $g=1=e$. Thus completing the proof.

Is this sufficient? I feel like I'm missing something crucial.

I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide

Kevin_H
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    You've got the right idea by appealing to the fact that $\gcd(a,b)=1$. However, you bring up various facts, and then don't use them. For example, you don't use your set $A$ or $B$ (and I don't really see how you would). Eventually, you conclude "since 10 and 21 are relatively prime, then $\langle a \rangle \cap \langle b \rangle = {1}$", but where did this come from? It doesn't seem to be built on your previous propositions. In so many words: good ideas, but I think your proof is structurally lacking. – Alex Wertheim Mar 04 '15 at 00:55

2 Answers2

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$\langle a \rangle \cap \langle b \rangle$ is a subgroup of $\langle a \rangle$ and of $\langle b \rangle$. By Lagrange the order of $\langle a \rangle \cap \langle b\rangle $ divides $10$ and $21$.

Asinomás
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Let $x \in \langle a \rangle \cap \langle b \rangle$. Then $e=x^{21}=(x^{10})^2\cdot x=e \cdot x = x$.

Here we have used that $x^{21}=e$ because $x \in \langle b \rangle$ and that $x^{10}=e$ because $x \in \langle a \rangle$.

The same argument works in general:

If $A$ and $B$ are subgroups of $G$ and have coprime order, then their intersection is trivial.

Indeed, let $x \in A \cap B$. Let $m=|A|$ and $n=|B|$. Write $1=rm+sn$. Then $x=x^1=x^{rm+sn}=(x^m)^r \cdot (x^n)^s=e \cdot e = e$.

lhf
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