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let $A$ and $B$ be two subgroups of $G$. we say that $B$ is a complement of $A$ if :

  1. $G=AB$

  2. $A\cap B=\{1\}$

Given a subgroup $A$ of $G$ i don't see how the complement $B$ of $A$ in $G$ is not unique, it seems to me like $A$ and $B$ partition $G$ right? I mean with these two conditions an element in $G$ must be lying in $A$ or in $B$, that is the subgoup $A$ has a complement if $(G-A)\cup\{1\}$ is also a subgroup of $G$ ?

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7  
Think about how many complements the $x$-axis has in $\mathbb{R}^2$. –  t.b. Jul 28 '11 at 9:36
4  
Note that AB is not the same as the union of A and B –  Tobias Kildetoft Jul 28 '11 at 9:52
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$A$ and $B$ cannot partition $G$, since they have an element in common. (It’s also true that their union won’t be $G$, but that’s less obvious.) –  Brian M. Scott Jul 28 '11 at 9:56

2 Answers 2

up vote 3 down vote accepted

Consider $G = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Take $A$ to be the cyclic subgroup generated by $(1,0)$, $B$ to be the cyclic subgroup generated by $(1,1)$, $C$ to be the cyclic subgroup generated by $(0,1)$. Then both $B$ and $C$ are complements of $A$ in $G$. You wrote "it seems to me like $A$ and $B$ partition $G$ right?", so perhaps you are thinking that every element of $G$ either belongs to $A$ or to $B$, but this is not true in general. In the example I just gave $(1,1)$ belongs to neither $A$ nor $C$, and $(0,1)$ belongs to neither $A$ nor $B$. Also consider the set $G - A \cup \{1\} = \{(0,0), (0,1), (1,1)\}$. This is certainly NOT a subgroup of $G$.

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1  
@palio: Try to see how this is related to Theo’s hint: algebra_fan’s $A$ corresponds to the $x$-axis in $\mathbb{R}$, his $B$ to the line $y=x$, and his $C$ to the $y$-axis. –  Brian M. Scott Jul 28 '11 at 10:01
    
@algebra_fan: excellent example, it explains it all!!! thanks. –  palio Jul 28 '11 at 10:06
    
@Brian M. Scott: yes i see that $(x,y)=(x,0)+(0,y)$ and $(x,y)=(x-y,0)+(y,y)$ –  palio Jul 28 '11 at 10:15
    
@palio: I am glad you've found my example helpful. You are very welcome! –  algebra_fan Jul 28 '11 at 10:31

I point out a couple of observations which have not yet been made. If $G$ is a finite group, and $G = AB$ for subgroups $A$ and $B$ with $A \cap B= 1$, then we have $|G| = |A||B|$ (see, for example, Herstein's ``Topics in Algebra"), which is usually much bigger than $|A| + |B| -1= |A \cup B|$. If any group $H$ (finite or not) has a factorization of the form $H = CD$, then we have $H = C^{h}D$ for any $h \in H$, since we can write $h = cd$ for some $c \in C, d \in D$. Then $C^{h}D = C^{cd}D = C^{d}D = (CD)^{d} = H^{d} = H.$ If we also had $C \cap D = 1$, then $C^{h} \cap D = C^{cd} \cap D = C^{d} \cap D = (C \cap D)^{d} = 1$. Thus the complements to $D$ are closed under conjugation, so the only chance for a complement to $D$ to be unique would be if it was normal. As shown by examples from the comments, even when the complement is normal, it need not be unique.

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For this reason, group theorists tend to count conjugacy classes of complements, rather than the complements themselves. When A is normal and abelian, then the set of conjugacy classes forms a group called the first cohomology, $H^1(B,A)$. In particular, in algebra_fan's example, $H^1(B,A) \cong A$ has two elements. Counting complements is also sometimes called counting crossed homomorphisms. –  Jack Schmidt Jul 28 '11 at 15:15

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