User gabe conant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:11:03Z http://mathoverflow.net/feeds/user/21240 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5528/when-does-a-subgroup-h-of-a-group-g-have-a-complement-in-g/87912#87912 Answer by Gabe Conant for When does a subgroup H of a group G have a complement in G? Gabe Conant 2012-02-08T17:52:30Z 2012-02-08T17:52:30Z <p>A lot can be said in the finitely generated abelian case, just by using the structure theorem. </p> <p>Call a group <strong>transversal</strong> if every subgroup has a complement; <strong>non-transversal</strong> of it has proper, nontrivial subgroups, none of which has a complement; and <strong>semi-transversal</strong> if it is not transversal but some proper, nontrivial subgroup has a complement.</p> <p>Take a finite abelian group $G=\mathbb{Z}/p_1^{e_1}\mathbb{Z}\times\ldots\times\mathbb{Z}/p_k^{e_k}\mathbb{Z}$ with $r\geq 0$, distinct primes $p_1,\ldots,p_k$ and $e_i\geq 1$. </p> <p>Then $G$ is transversal when $e_i=1$ for all $i$, semi-transversal when $k>1$ and $e_i>1$ for some $i$, and non-transversal when $k=1$ and $e_1>1$.</p> <p>For an infinite finitely generated abelian group $G$, $G$ is non-transversal if $G=\mathbb{Z}$, and semi-transversal otherwise.</p> <p>Given a finitely generated abelian group $G$ and $S\subseteq G$, call $S$ <strong>independent</strong> if $0\not\in S$ and for all $x_1,\ldots,x_k\in S$, $r_1,\ldots, r_k\in\mathbb{Z}$, we have that $\sum_{i=1}^k r_ix_i=0$ implies $r_ix_i=0$ for all $i$. Call $S$ a <strong>basis</strong> of $G$ if it is independent and $G=\left&lt; S\right>$.</p> <p>Then if $H\leq G$, $H$ has a complement if and only if $H$ has a basis that can be expanded to a basis of $G$. If $S$ is a basis for $H$ and $S\subseteq T$ for some basis $T$ of $G$, then the complement of $H$ is $\left&lt; T\backslash S\right>$.</p> <p>So if $G$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$ then any subgroup has a complement, since any subspace has a basis that can be completed to the whole space. My definition of independence is the same as linear independence.</p> <p>If $G$ is a free module over $\mathbb{Z}/p^n\mathbb{Z}$ then my definition of independence is weaker than linear independence. I would like to say that $H\leq G$ has a complement if and only if it has a module basis, but I can't prove the reverse direction of this.</p> <p>I know less about the divisible abelian case, except that if $G$ is a divisible abelian group then saying that $H\leq G$ has a complement is the same as saying that $H$ is divisible. In particular $\mathbb{Q}$ and the Prufer $p$-group $\mathbb{Z}(p^\infty)$ are both non-transversal.</p>