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Why does the subgroup have to be able divide the group? For example why isn't the group:
$S= \{-4,-3,-1,0,1,2,3,4\}$
a subgroup of $G= \{-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 \}$? Aren't both additive, have a neutral number, each member has a reciprocal and every member in group $S$ is present in group $G$?

Why do subgroups have such a specific requirement of being be able to divide the group?

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I'm confused as to what these 'groups' you've denoted are supposed to be. If you add -6 to -6, you get -12, which is not in $G$, so $G$ is not a group as I understand it. –  Alastair Litterick Apr 24 '12 at 13:25
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I'd also like to know where you found the phrase "divide the group", as I think you've misunderstood something. Lagrange's Theorem tells us that the order of a subgroup divides the order of the group, but we prove this after defining subgroups. –  Alastair Litterick Apr 24 '12 at 13:33
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As mentioned by Alastair Litterick, I think you are looking for this: en.wikipedia.org/wiki/Lagrange%27s_theorem_%28group_theory%29 but you should probably try to understand the basic definitions of groups, subgroups and cosets before you try to understand a proof of Lagrange's theorem. –  Sam Jones Apr 24 '12 at 13:38
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Here closureness properties are not satisfied so i dont think they are groups under addition. Here $2+3=5\not\in S$ and $3+4=7\not\in G$. –  Kns Apr 24 '12 at 16:09

2 Answers 2

up vote 5 down vote accepted

A group must be closed under the group operation. Since you said "both are additive," I am going to assume that the group operation on $G$ (and hence $S$) is addition. Then $G$ is not closed under addition ($6 + 1 = 7 \notin G$, for example), unless you are considering $G$ as $\mathbb{Z}/13$. The same issue happens with $S$: $3 + 1 = 4 \notin S$, so $S$ is not closed under the group operation and is hence not a subgroup.

Also, the additive inverse of $-4 \in S$ is not in $S$.

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If $G$ is a finite group, then according to Lagrange's theorem the order (number of elements) of a subgroup $H$ divides the order of $G$. A proof of this fact can be found in any introductory text on abstract algebra.

The basic idea is that $G$ can be split up into pairwise disjoint parts called cosets that each have the same order $H$ does. Therefore $G$ has order $n \cdot |H|$, where $n$ is the number of these parts and $|H|$ the order of $H$.

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And just to clarify, here "pairwise disjoint" means that two different parts have no elements in common. –  Mikko Korhonen Apr 24 '12 at 14:05

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