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I want to show that the order of $ab$ is the same as the product of orders of $a$ and $b.$

So at title, $|ab| = |a||b|.$

Here $|a|$ means the order of an element $a$ in a group $G$. So $a^n = e$ if the order of $a$ is n, and $e$ is an identity of a group.

I think it's false.

Let $G$ be a group of $\Bbb Z$ mod $4$ under addition modulo $n$. The $G = \{0,1,2,3\} .$ Let $a = 2$ and $b = 3$. Then the order of $a$ is $2,$ and that of $b$ is $4$.

However the order of $ab$ is $4$.

So $|ab|$ is not equal to $|a||b|.$

Is this correct thought?

Shaun
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jun
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    Your thinking is correct. In fact, even better example would be an element $a\ne e$ and $b=a^{-1}$. In that case, $|a|=|b|>1$ but $|ab|=|e|=1$. –  Sep 04 '20 at 16:27
  • but can I say 6 is an element of my group G? I said G = {0,1, 2,3}. 6 mod 4 is 2 so 6 is in a group? – jun Sep 04 '20 at 16:28
  • $6 \mod 4 = 2 \in G$ – amWhy Sep 04 '20 at 16:30
  • In general we cannot say much on the order of a product in a group, see this post, or this post. – Dietrich Burde Sep 04 '20 at 16:33
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    @jun As you may know, it doesn't really matter what the elements of the group are as long as they satisfy the same "multiplication table". Strictly speaking, group of integers $\pmod 4$ is not equal to ${0,1,2,3}$ but is equal to ${0_{\equiv 4}, 1_{\equiv 4}, 2_{\equiv 4}, 3_{\equiv 4}}$, i.e. its elements are equivalence classes of integers where the equivalence relation is $\equiv\pmod 4$. (In other words, $G={{\ldots,-8,-4,0,4,8,\ldots},{\ldots,-7,-3,1,5,9,\ldots},{\ldots,-6,-2,2,6,10,\ldots},{\ldots,-5,-1,3,7,11,\ldots}}$ as a set. –  Sep 04 '20 at 16:39
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    (Cont'd) However, you can instead take any four elements and pretend $G$ has those elements, as long as you keep the rules of addition in this group the same. The bottom line is, don't stress too much about the actual nature of elements of your group as long as you know the rules of operation. –  Sep 04 '20 at 16:40
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    To go a little above and beyond, while the proof is somewhat more complicated, you can show that for any $m$, $n$ and $p$ greater than $1$ (including 'infinite'), there's a group that has elements $a$ of order $m$ and $b$ of order $n$ such that $ab$ has order $p$ (where an element $g$ having infinite order means there's no $i$ such that $g^i$ is the identity). – Steven Stadnicki Sep 04 '20 at 16:46

2 Answers2

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Such a relation $o(ab)=o(a)o(b)$ does not hold in general, not even in finite abelian groups, where we have $o(ab)\mid lcm(o(a),o(b))$. You have given a counterexample, with $lcm(2,4)=4\neq 8$.

References:

Order of products of elements in a finite Abelian group

Possible order of $ab$ when orders of $a$ and $b$ are known.

Dietrich Burde
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    It's not true that in a finite abelian group we have $o(ab)=lcm(o(a),o(b))$, e.g. choose $a$ to be non-trivial and $b = a^{-1}$. In a finite abelian group it is true that $lcm(o(a),o(b)) = lcm(o(a),o(ab)) = lcm(o(b),o(ab))$, and this is the best result possible in the sense that if $m$, $n$ and $k$ are positive integers such that $lcm(m,n) = lcm(m,k) = lcm(n,k)$ then there is a finite abelian group with elements $a$ and $b$ such that $o(a) = m$, $o(b) = n$ and $o(ab) = k$. – Zoe H Sep 05 '20 at 03:53
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    @ZoeH Sorry, I wanted to write that $o(ab)$ divides the lcm, this was a typo (see the first link). – Dietrich Burde Sep 05 '20 at 08:14
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Yes, you're mostly correct; only: in

Let $G$ be a group of $\Bbb Z$ mod $4$ under addition modulo $n$.

your $n$ should be $4$, and indeed $ab=2+3=5\equiv 1\pmod{4}$ has order $4$, so the counterexample holds.

Shaun
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