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Let $G$ be an abelian transitive permutation group acting on $\Omega$. Let $\{g_1, \ldots, g_n\}$ be elements of $G$ such that, $\forall i \not = j$, there is no $k \in \mathbb{N}$ such that $g_i = g_j^k$ or $g_j = g_i^k$.

Question: Is it true that, $\displaystyle g_1^{x_1} = \prod_{i=2}^n g_i^{x_i}$ implies $g_1^{x_1} = 1$ (for $x \in \mathbb{N}^n$)?

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This is a very simple problem. Your condition ensures that $g_1,g_2,\ldots,g_n$ generate different cyclic subgroups, but it could be that $g_1 \in \langle g_2 , g_3 \rangle$, for example. An explicit example where this happens is when $G$ is the normal Klein 4-subgroup of the symmetric group $S_4$ and $\Omega = \{1,2,3,4 \}.$ So the answer is "not in general". –  Geoff Robinson Aug 17 '11 at 13:27
    
But $S_4$ isn't an abelian transitive group, no? –  Zac Aug 17 '11 at 14:14
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He never said it was! –  Derek Holt Aug 17 '11 at 14:20
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Let $g_1 = (12)(34)$, $g_2 = (14)(23)$ and $g_3 = (13)(24)$. We have $g_{1}^{1} = g_{2}^{1} g_{3}^{1}$, but $g_{1}^{1} \neq 1.$ –  Geoff Robinson Aug 17 '11 at 14:48
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Cyclic groups of order $pq$ for distinct primes $p$ and $q$ seem to work. –  Geoff Robinson Aug 17 '11 at 22:04
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1 Answer

g_1=(1,1), g_2 = (1,0), g_3=(0,1)

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The MathOverflot bot keeps bumping this question to the front page, so I'll vote this up to stop that, but I wouldn't otherwise. Here are the details MrT left out: $G = \mathbb{Z}\times \mathbb{Z}$ and the group operation is componentwise addition. So you need to translate the question into additive notation to see that this answer is right. Clearly $g_1$ is not a power of either of the others because $(1,0)^n = (n,0)$ and similarly for $(0,1)$. Next, $(1,1) = \prod g_i$ because $(1,1)=(0,1)+(1,0)$. But $(1,1)$ is not the "one" in the group because it's not the additive identity. –  David White Sep 28 '11 at 16:51
    
$\mathbb Z\times \mathbb Z$ is not a permutation group. The Klein Four group works, though. –  Will Sawin Sep 28 '11 at 18:04
    
Ah, I missed that he wanted a permutation group. Oh well, I already voted for this answer and now it's too late to take back. At least this question will not reappear on the front-page –  David White Sep 29 '11 at 14:11
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