# uniqueness of a complement of a subgroup

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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Think about how many complements the $x$-axis has in $\mathbb{R}^2$. – t.b. Jul 28 '11 at 9:36
Note that AB is not the same as the union of A and B – Tobias Kildetoft Jul 28 '11 at 9:52
$A$ 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

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 $G$.
@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
@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
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.
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