- #1
logan3
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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 $H \oplus K = m + n$.
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 $H \oplus K = m + n$.
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