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I am trying to prove (by induction on k) that if $o(a_i) = n_i$ for all $i \in \mathbb{N}$ and $gcd(n_i,n_j)=1$ for all $i,j \in \mathbb{N}$ then $o(a_1a_2\dots a_k) = n_1n_2\dots n_k$. Where $o(n)$ denotes the order of the element $n$.

My base case is if $k=2$ then $o(a_1a_2) = n_1n_2$. This result was proven in an earlier exercise.

I let my inductive hypothesis be: $o(a_1a_2\dots a_m) = n_1n_2\dots n_m$ for all $1 \leq m \leq k$. However, I'm having trouble showing how:

$o(a_1a_2\dots a_ka_{k+1}) = n_1n_2\dots n_kn_{k+1}$.

I can start with $o(a_1a_2\dots a_k)o(a_{k+1}) = (n_1n_2\dots n_k)(n_{k+1}) = n_1n_2\dots n_kn_{k+1}$. But then I would have to prove that $o(a_1a_2\dots a_{k+1}) = o(a_1a_2\dots a_k)o(a_{k+1}).$

Would it be okay to to group $n_k$ and $n_{k+1}$ as one term so that I technically get $k$ terms instead of $k+1$ terms so that: $o(a_1a_2\dots a_k)o(a_{k+1}) = (n_1n_2\dots n_k)(n_{k+1}) = (n_1n_2\dots n_{k-1})(n_kn_{k+1}) = o(a_1 \dots (a_ka_{k+1}))$ by the inductive hypothesis?

All of this seems a bit pedantic, so I would like a little guidance on what would be good convention.

meijiu
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  • What exactly is the setting here? Are the $a_i$ elements of a group? Is the group abelian? Are they elements in a ring? Is then the order necessarily defined? – Arthur Sep 19 '17 at 13:08
  • Sorry for not including this important information, all $a_i$ are in an abelian group G – meijiu Sep 19 '17 at 13:12

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Once you've proven that it is true for $k = 1$ (i.e. for a product of just two elements), then the rest follows by induction. The equality $o(a_1a_2\cdots a_{k+1}) = o(a_1a_2\cdots a_k)o(a_{k+1})$ comes from the base case: Note that $a_1a_2\cdots a_k$ is some element with some order (we just happen to know a "factorisation" of it, and choose to write it that way, but make no mistake: it is a single element), and $a_{k+1}$ is an element with some other coprime order. Therefore the order of the product of these two elements is the products of their respective orders.

Arthur
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