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Given $G$ is a group, $a \in G$, $o(a) = 218$, compute $o(a^{63})$

Well, first I started by saying that $gcd(218,63) = 1$. Thus,

$218 | 63k$, meaning $63k = 218s$ and $k,s \in Z$

$o(a) = 218 \rightarrow a^{218} = e$

Now I need to find such $k$ that will give me $e$ by doing that: $a^{63k} = e$

I'm not sure about the value of $k$ yet, can anyone help please. (Obviously it can be $k = 218$ I guess).

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    You don't seem to be using the most important thing you have included: The fact that the numbers are coprime. Why did you include this and then not use it? – 2017-01-10
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    @TobiasKildetoft I did try with Bezout Lemma, I believe there need to be used Euclid Lemma here. – 2017-01-10
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    You are looking for the smallest value of $k$ such that $63k$ is divisible by $218$. – 2017-01-10

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Let $u,v \in \mathbb Z$ such that $218u+63v=1$, given by Bézout's lemma.

Then $a = a^1 = a^{218u} a^{63v} = (a^{63})^v$.

Therefore, the subgroup generated by $a^{63}$ contains $a$ and so coincides with the subgroup generated by $a$. This implies that $a^{63}$ has the same order as $a$.

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    I see, so I had to use this: http://math.stackexchange.com/questions/2089141/given-g-group-and-a-in-g-an-e-gcdm-n-1-prove-exists-b-in Thank you sir. – 2017-01-10