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Every group of order $231$ is the direct product of a group of order $11$ and a group of order $21$.

By Sylow's theorem, we know there are one Sylow-7 subgroup, one Sylow-11 subgroup (these two are normal, for sure), and some Sylow-2 subgroups.

What does the "direct product" mean in this context? Thanks.

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@QiaochuYuan In that wiki page, $|G\times H|=|G||H|$, which seems not true in this context. :) –  Vladimir Oct 7 '12 at 2:11
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I don't follow. The last time I checked, $231 = 11 \cdot 21$. –  Qiaochu Yuan Oct 7 '12 at 2:20
    
@QiaochuYuan Hmmm... I was writing it as a "direct product" of a Sylow-7, a Sylow-11 and some Sylow-3 subgroups, whose intersection with each other are all trivial. So in fact, this is not a direct product... Could you show me how to deal with it? –  Vladimir Oct 7 '12 at 2:30
    
The Sylow $3$-subgroup and the Sylow $7$-subgroup need not commute with each other in general. –  Qiaochu Yuan Oct 7 '12 at 3:17
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2 Answers 2

It means there is a group $\langle G, \star\rangle $ of order 11, and a group $\langle H, \ast\rangle$ of order 21, so that your group of order 231 has elements of the form $(g, h)$ where $g\in G$ and $h\in H$, and a group operation that operates componentwise, so that $(g,h)\cdot(g', h') = (g\star g', h\ast h')$, and $(g, h)^{-1} = (g^{-1}, h^{-1})$.

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It means that there exist two normal subgroups $K_1$ and $K_2$ such that $K_1\cap K_2=\{e\}$ and every element in the group can be written as $k_1k_2$, with $k_1\in K_1$, $k_2\in K_2$.

This is equivalent to saying that the group is isomorphic to $K_1\times K_2$ with the entrywise operation.

(if you are familiar with the direct sum, it is exactly the same idea but with the operation written multiplicatively).

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i.e. $K$ is an internal direct product of $K_1$ and $K_2$ –  Kris Oct 6 '12 at 22:49
    
Thanks. So in this context, it suffices to see if the intersection of those two subgroups is trivial, right? –  Vladimir Oct 7 '12 at 0:33
    
Yes, because then the product will have exactly 231 elements and it will have to be the whole group. –  Martin Argerami Oct 7 '12 at 1:00
    
Note that the subgroups $K_1$ and $K_2$ have to be normal. –  Mikko Korhonen Oct 12 '12 at 8:34
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