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Let $G_1,G_2,...,G_n$ be groups. Show that the order of an element $(a_1,a_2,...,a_n)$ $\in$ $G_1 \times G_2 \times\cdots\times G_n$ is lcm($o(a_1),...,o(a_n))$.

I know I need to use the fact that the least common multiple of positive integers $x_1,x_2,...,x_n$ is the unique positive multiple of $x_1,x_2,...,x_n$ that divides all other such multiples.

Note on notation: for the case where $o(a_i)=\aleph_0$ one defines lcm($o(a_1),...,o(a_n))=\aleph_0$

I was looking at this question but I'm not sure if I can assume the groups are abelian in my case.

Any advice would be greatly appreciated! Thanks.

José Carlos Santos
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user2553807
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    Fortunately, even if the groups are not abelian, anything in any of the factors will commute with anything in a different factor. So the same type of argument applies. – Tobias Kildetoft Feb 20 '14 at 14:16

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The best way I know to do this is by induction, it's one of those nice cases where $n=1$ isn't saying anything, and the inductive step for general $n$ is really just the inductive step for $n=2$, since $G_1\times \dots \times G_n=(G_1\times \dots \times G_{n-1})\times G_n$ and:

lcm$(a_1,\dots,a_n)=$ lcm$($lcm$(a_1,\dots,a_{n-1}),a_n))$

So, can you show that $o(g_1,g_2) = lcm(o(g_1),o(g_2))$? Clearly if one of the $g_i$ has infinite order, then so does $(g_1,g_2)$, so that just leaves the finite case. Remember that if the order of an element $g$ is $d$, then $g^n=1 \iff d|n$.

Tom Oldfield
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