Group Theory: Proving Finite Rank of Torsion-free Abelian Groups Using Independence

[logan3]

If someone can check this, it would be appreciated. (Maybe it can submitted for a POTW afterwards.) Thank-you.

PROBLEM
Prove that if $H$ and $K$ are torsion-free groups of finite rank $m$ and $n$ respectively, then $G = H \oplus K$ is of rank $m + n$.

SOLUTION
Let $h_1, ..., h_m$ and $k_1, ..., k_n$ be sets of independent elements of $H$ and $K$, respectively. Then the set $h_1, ..., h_m, k_1, ..., k_n$ is also independent.

If $r_1 h_1 + \cdots + r_m h_m + s_1 k_1 + \cdots + s_n k_n = 0$, then $r_1 h_1 + \cdots + r_m h_m = -s_1 k_1 - \cdots - s_n k_n$. But since $H \cap K = {0}$, then $r_1 h_1 + \cdots + r_m h_m = -s_1 k_1 - \cdots - s_n k_n = 0$. Therefore, by independence of $h_1, ..., h_m$ and $k_1, ..., k_n$ $r_1 = r_2 = \cdots = r_m = s_1 = s_2 = \cdots = s_n = 0$.

Next, suppose ${h_1, ..., h_m, k_1, ..., k_n}$ is not maximal and there exists an element $h + k$, where $h \in H$ and $k \in K$, s.t. ${h_1, ..., h_m, k_1, ..., k_n, h + k}$ is independent. Since ${h_1, ..., h_m,h}$ is not independent, there exists a set of integers $t_1, ..., t_m, t$ that are not all zero, s.t. $t_1 h_1 + \cdots + t_m h_m + th = 0$. Thus if $t = 0$, then ${h_1, ..., h_m}$ cannot be independent, because at least one of $t_1, ..., t_m$ is nonzero. Therefore, $t \ne 0$.

Next, because ${k, k_1, ..., k_n}$ is not independent (i.e. there exist $s, s_1, ..., s_n$, not all zero, s.t. $sk + s_1 k_1 + \cdots + s_n k_n = 0$) and using the same argument as above, it follows that $s \ne 0$. Thus,

$st_1 h_1 + \cdots + st_m h_m + st(h + k) + ts_1 k_1 + \cdots + ts_n k_n = s(th + t_1 h_1 + \cdots + t_m h_m) + t(sk + s_1 k_1 + \cdots + s_n k_n) = 0$

But since $st \ne 0$, then ${h_1, h_2, ..., h_m, k_1, ..., k_n, h + k}$ is not independent. Therefore, ${h_1, h_2, ..., h_m, k_1, ..., k_n}$ is a maximal independent set and rank $H \oplus K = m + n$.

 

[Euge]

Great work, Logan! Your proof is correct. The question would not be submitted as a POTW, though. If you want to propose a POTW, then click on the blue POTW tab above, then click on the link under "this form" under the last section. Your proposal may or may not be accepted, but if it is, then it will be noted for an upcoming POTW.