Question in Algebra by Serge Lang.
$begingroup$
The following is a lemma in Algebra (page 44):
$A$ is a finite abelian p-group. Let $overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a
representative a of $overline{b}$ in $A$ which also has period $p^r$.
Proof. Let b be any representative of $overline{b}$ in A. Then $p^rb$ lies in A, say
$p^rb = na$ with some integer $n ge 0$. We note that the period of $overline{b}$ is $le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^kmu$, where $mu$ is prime to $p$. Then $mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k le r_1$. Then $p^kmu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period
$$p^{r+r_1-k}$$
whence by hypothesis,$underline{r + r_1 - k le r_1}$ and $r le k$. This proves that there exists an element $cin A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .
My question is, how the underline part in the proof makes sense.
abstract-algebra torsion-groups
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
$endgroup$
add a comment |
$begingroup$
The following is a lemma in Algebra (page 44):
$A$ is a finite abelian p-group. Let $overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a
representative a of $overline{b}$ in $A$ which also has period $p^r$.
Proof. Let b be any representative of $overline{b}$ in A. Then $p^rb$ lies in A, say
$p^rb = na$ with some integer $n ge 0$. We note that the period of $overline{b}$ is $le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^kmu$, where $mu$ is prime to $p$. Then $mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k le r_1$. Then $p^kmu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period
$$p^{r+r_1-k}$$
whence by hypothesis,$underline{r + r_1 - k le r_1}$ and $r le k$. This proves that there exists an element $cin A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .
My question is, how the underline part in the proof makes sense.
abstract-algebra torsion-groups
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
$endgroup$
1
$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49
add a comment |
$begingroup$
The following is a lemma in Algebra (page 44):
$A$ is a finite abelian p-group. Let $overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a
representative a of $overline{b}$ in $A$ which also has period $p^r$.
Proof. Let b be any representative of $overline{b}$ in A. Then $p^rb$ lies in A, say
$p^rb = na$ with some integer $n ge 0$. We note that the period of $overline{b}$ is $le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^kmu$, where $mu$ is prime to $p$. Then $mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k le r_1$. Then $p^kmu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period
$$p^{r+r_1-k}$$
whence by hypothesis,$underline{r + r_1 - k le r_1}$ and $r le k$. This proves that there exists an element $cin A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .
My question is, how the underline part in the proof makes sense.
abstract-algebra torsion-groups
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
$endgroup$
The following is a lemma in Algebra (page 44):
$A$ is a finite abelian p-group. Let $overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a
representative a of $overline{b}$ in $A$ which also has period $p^r$.
Proof. Let b be any representative of $overline{b}$ in A. Then $p^rb$ lies in A, say
$p^rb = na$ with some integer $n ge 0$. We note that the period of $overline{b}$ is $le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^kmu$, where $mu$ is prime to $p$. Then $mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k le r_1$. Then $p^kmu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period
$$p^{r+r_1-k}$$
whence by hypothesis,$underline{r + r_1 - k le r_1}$ and $r le k$. This proves that there exists an element $cin A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .
My question is, how the underline part in the proof makes sense.
abstract-algebra torsion-groups
abstract-algebra torsion-groups
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
edited Dec 11 '18 at 10:48
X.T Chen
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
asked Dec 11 '18 at 10:28
X.T ChenX.T Chen
918
918
1
$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49
add a comment |
1
$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49
1
1
$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49
add a comment |
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$begingroup$
What is $;A;$, though?
$endgroup$
– DonAntonio
Dec 11 '18 at 10:34
$begingroup$
@DonAntonio I have edited
$endgroup$
– X.T Chen
Dec 11 '18 at 10:49