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Let $(a_1,a_2,...,a_n)\in G_1\oplus G_2 \oplus \cdot \cdot \cdot \oplus G_n$. Give a necessary and sufficient condition for $|(a_1,a_2,...,a_n)| = \infty$

I know $|(a_1,a_2,...,a_n)|$ is related to the LCM somehow, but I'm confused on the context of the question.

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The $G_i$ are groups, I suppose? And $|\cdot|$ denotes the order of an element? Hint: If $a_i$ has finite order $o_i$ for each $i$, what can you say about $(a_1,\ldots, a_n)^{o_1\cdots o_n}$? – martini May 3 '12 at 6:44
up vote 1 down vote accepted

For a (multiplicative) group $G$ and $g\in G$, set $|g|=m\gt 0$ if $g^m=1$ and $g^k\neq 1$ for all $k$, $0\lt k\lt m$; and $|g|=0$ if $g$ is not a torsion element.

If $G_1,\ldots,G_n$ are a finite collection of groups, then $$|(g_1,\ldots,g_n)| = \mathrm{lcm}(|g_1|,|g_2|,\ldots,|g_n|).$$ (Prove it; remember that $\mathrm{lcm}(0,n) = 0$ for all $n$).

More generally, if $G$ is a group, $g_1$ and $g_2$ are two elements that commute, and $\langle g_1\rangle\cap\langle g_2\rangle = \{1\}$, then $|g_1g_2| = \mathrm{lcm}(|g_1|,|g_2|)$: since $g_1$ and $g_2$ commute, then $(g_1g_2)^n = g_1^ng_2^n$. If $(g_1g_2)^k = 1$, then $g_1^kg_2^k=1$, so $g_1^k = g_2^{-k}\in\langle g_1\rangle\cap\langle g_2\rangle=\{1\}$, so $g_1^k = g_2^k = 1$, hence $|g_1|$ divides $k$, and $|g_2|$ divides $k$.

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I think the required answer is that $(g_1, g_2, ..., g_n)$ has infinite order if and only it at least one such $g_i$ has infinite order in the corresponding $G_i$.

If each $g_i$ had finite order in the corresponding $G_i$ then $(g_1, g_2, ..., g_n)$ would have finite order in $G$ (have a think about this).

Conversely if the order of $(g_1, ..., g_n)$ is finite then each $g_i$ has a (finite) power giving the identity in the corresponding $G_i$. This tells you that all $g_i$'s have finite order.

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Set $a=(a_1,\cdots,a_n)$ to be an arbitrary element of $H=\bigoplus_i G_i$ and $e=(e_1,\cdots,e_n)$ the identity. Note that each $e_i$ is the identity of the direct summand $G_i$. Now some thoughts to ponder:

  • $a^m=e$ is true if and only if $a_i^m=e_i$ for each $i=1,2,\cdots,n$.
  • $a_i^m=e_i$ is true if and only if the order of $a_i$ (in $G_i$) divides $m$ or $a_i=e_i$.
  • If the order of every $a_i$ divides $m$, then $a_i^m=e_i$ for each $i$.
  • An element of $a\in H$ for which $a^m=e$ for some $m\in\Bbb Z$ must have finite order. (Why?)
  • There is an $m\in\Bbb Z$ that is divisible by each $a_i$'s order. (Multiply, take the LCM, etc.)

If each $a_i$ has finite order, the above reasoning tells us $a$ has finite order too. Now, suppose at least one of the $a_i$'s has infinite order. Can $a\in H$ have finite order? (Assume it does and see what happens...)

Finally, there are precisely $n$ conditions on the table: whether or not $a_i$ has finite order, separately for each $i$. What is sufficient for $a$ to have infinite order, in terms of the $a_i$'s orders? And then what is necessary for $a$ to have infinite order? So then what can we say is both necessary and sufficient?

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