Theorem on abelian groups when the factor is free
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Lemma: Suppose that factor group $A/B=bigoplus limits_{i=1}^{n}(A_i/B)$ - direct sum, where $B$ - direct addend in
each subgroup $A_i$, i.e. $A_i=Boplus J_i$. Then
$$A=Boplusleft(bigopluslimits_{i=1}^{n}J_i right).$$
Theorem: Suppose that $A$ - abelian group, $B$ - subgroup of $A$. If the factor group $A/B$ is free then $A$ is direct sum of $B$ and
free group $F^{ab}$, i.e. $A=Boplus F^{ab}.$
Proof: Since $A/B$ - direct sum of infinite cyclic groups, then by lemma it is enough to consider the case when $A/Bcong
mathbb{Z}$. Thus, $A/B=langle bar{a}rangle cong mathbb{Z}.$
Taking $0neq a_0in bar{a}$ (element of coset $bar{a}$, which does
not lie in $B$). Then elements $ka_0$ be representatives of cosets
$kbar{a}$, $k=0,pm 1,pm 2,dots,$ i.e. $A=Boplus langle a_0rangle
$.
Let me ask you some questions which I was not able to answer by myself, please.
1) I spent some time in order to understand why it suffices to consider the case when $A/Bcong mathbb{Z}$? How the lemma 4 is applied here? I mean since $A/B$ is free then $A/B=bigopluslimits_{i=1}^{n}a_imathbb{Z}$ but I am not able to understand what is $A_i, B$ and $J_i$ here with respect to the notations of lemma ?
2) I have shown that $A=B+langle a_0rangle$. But how to show that the sum is direct? In other words, $Bcap langle a_0rangle={0}$? Suppose that $x$ lies in the intersection then $x=ka_0=b$ but I don't know what to do next?
I would be very grateful for help! I have spent some hours trying to write down everything accurately but nothing.
EDIT: Possible answer to question 2. We have shown that $A=B+langle a_0rangle$. In order to show that the sum is direct we have to show that $Bcaplangle a_0rangle={0}$. Suppose $xin Bcaplangle a_0rangle$ then $x=b=ka_0$ and since $a_0in bar{a}=a+B$ then $a_0=a+b'$ where $b'in B$ then from $b=ka_0$ we have $b=ka+kb'$ hence $b'':=b-kb'=ka$ i.e. $kain B$ but since $bar{a}=a+B$ has order infinite order then $k=0$. Thus $x=0$. Right?
group-theory
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Lemma: Suppose that factor group $A/B=bigoplus limits_{i=1}^{n}(A_i/B)$ - direct sum, where $B$ - direct addend in
each subgroup $A_i$, i.e. $A_i=Boplus J_i$. Then
$$A=Boplusleft(bigopluslimits_{i=1}^{n}J_i right).$$
Theorem: Suppose that $A$ - abelian group, $B$ - subgroup of $A$. If the factor group $A/B$ is free then $A$ is direct sum of $B$ and
free group $F^{ab}$, i.e. $A=Boplus F^{ab}.$
Proof: Since $A/B$ - direct sum of infinite cyclic groups, then by lemma it is enough to consider the case when $A/Bcong
mathbb{Z}$. Thus, $A/B=langle bar{a}rangle cong mathbb{Z}.$
Taking $0neq a_0in bar{a}$ (element of coset $bar{a}$, which does
not lie in $B$). Then elements $ka_0$ be representatives of cosets
$kbar{a}$, $k=0,pm 1,pm 2,dots,$ i.e. $A=Boplus langle a_0rangle
$.
Let me ask you some questions which I was not able to answer by myself, please.
1) I spent some time in order to understand why it suffices to consider the case when $A/Bcong mathbb{Z}$? How the lemma 4 is applied here? I mean since $A/B$ is free then $A/B=bigopluslimits_{i=1}^{n}a_imathbb{Z}$ but I am not able to understand what is $A_i, B$ and $J_i$ here with respect to the notations of lemma ?
2) I have shown that $A=B+langle a_0rangle$. But how to show that the sum is direct? In other words, $Bcap langle a_0rangle={0}$? Suppose that $x$ lies in the intersection then $x=ka_0=b$ but I don't know what to do next?
I would be very grateful for help! I have spent some hours trying to write down everything accurately but nothing.
EDIT: Possible answer to question 2. We have shown that $A=B+langle a_0rangle$. In order to show that the sum is direct we have to show that $Bcaplangle a_0rangle={0}$. Suppose $xin Bcaplangle a_0rangle$ then $x=b=ka_0$ and since $a_0in bar{a}=a+B$ then $a_0=a+b'$ where $b'in B$ then from $b=ka_0$ we have $b=ka+kb'$ hence $b'':=b-kb'=ka$ i.e. $kain B$ but since $bar{a}=a+B$ has order infinite order then $k=0$. Thus $x=0$. Right?
group-theory
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Lemma: Suppose that factor group $A/B=bigoplus limits_{i=1}^{n}(A_i/B)$ - direct sum, where $B$ - direct addend in
each subgroup $A_i$, i.e. $A_i=Boplus J_i$. Then
$$A=Boplusleft(bigopluslimits_{i=1}^{n}J_i right).$$
Theorem: Suppose that $A$ - abelian group, $B$ - subgroup of $A$. If the factor group $A/B$ is free then $A$ is direct sum of $B$ and
free group $F^{ab}$, i.e. $A=Boplus F^{ab}.$
Proof: Since $A/B$ - direct sum of infinite cyclic groups, then by lemma it is enough to consider the case when $A/Bcong
mathbb{Z}$. Thus, $A/B=langle bar{a}rangle cong mathbb{Z}.$
Taking $0neq a_0in bar{a}$ (element of coset $bar{a}$, which does
not lie in $B$). Then elements $ka_0$ be representatives of cosets
$kbar{a}$, $k=0,pm 1,pm 2,dots,$ i.e. $A=Boplus langle a_0rangle
$.
Let me ask you some questions which I was not able to answer by myself, please.
1) I spent some time in order to understand why it suffices to consider the case when $A/Bcong mathbb{Z}$? How the lemma 4 is applied here? I mean since $A/B$ is free then $A/B=bigopluslimits_{i=1}^{n}a_imathbb{Z}$ but I am not able to understand what is $A_i, B$ and $J_i$ here with respect to the notations of lemma ?
2) I have shown that $A=B+langle a_0rangle$. But how to show that the sum is direct? In other words, $Bcap langle a_0rangle={0}$? Suppose that $x$ lies in the intersection then $x=ka_0=b$ but I don't know what to do next?
I would be very grateful for help! I have spent some hours trying to write down everything accurately but nothing.
EDIT: Possible answer to question 2. We have shown that $A=B+langle a_0rangle$. In order to show that the sum is direct we have to show that $Bcaplangle a_0rangle={0}$. Suppose $xin Bcaplangle a_0rangle$ then $x=b=ka_0$ and since $a_0in bar{a}=a+B$ then $a_0=a+b'$ where $b'in B$ then from $b=ka_0$ we have $b=ka+kb'$ hence $b'':=b-kb'=ka$ i.e. $kain B$ but since $bar{a}=a+B$ has order infinite order then $k=0$. Thus $x=0$. Right?
group-theory
Lemma: Suppose that factor group $A/B=bigoplus limits_{i=1}^{n}(A_i/B)$ - direct sum, where $B$ - direct addend in
each subgroup $A_i$, i.e. $A_i=Boplus J_i$. Then
$$A=Boplusleft(bigopluslimits_{i=1}^{n}J_i right).$$
Theorem: Suppose that $A$ - abelian group, $B$ - subgroup of $A$. If the factor group $A/B$ is free then $A$ is direct sum of $B$ and
free group $F^{ab}$, i.e. $A=Boplus F^{ab}.$
Proof: Since $A/B$ - direct sum of infinite cyclic groups, then by lemma it is enough to consider the case when $A/Bcong
mathbb{Z}$. Thus, $A/B=langle bar{a}rangle cong mathbb{Z}.$
Taking $0neq a_0in bar{a}$ (element of coset $bar{a}$, which does
not lie in $B$). Then elements $ka_0$ be representatives of cosets
$kbar{a}$, $k=0,pm 1,pm 2,dots,$ i.e. $A=Boplus langle a_0rangle
$.
Let me ask you some questions which I was not able to answer by myself, please.
1) I spent some time in order to understand why it suffices to consider the case when $A/Bcong mathbb{Z}$? How the lemma 4 is applied here? I mean since $A/B$ is free then $A/B=bigopluslimits_{i=1}^{n}a_imathbb{Z}$ but I am not able to understand what is $A_i, B$ and $J_i$ here with respect to the notations of lemma ?
2) I have shown that $A=B+langle a_0rangle$. But how to show that the sum is direct? In other words, $Bcap langle a_0rangle={0}$? Suppose that $x$ lies in the intersection then $x=ka_0=b$ but I don't know what to do next?
I would be very grateful for help! I have spent some hours trying to write down everything accurately but nothing.
EDIT: Possible answer to question 2. We have shown that $A=B+langle a_0rangle$. In order to show that the sum is direct we have to show that $Bcaplangle a_0rangle={0}$. Suppose $xin Bcaplangle a_0rangle$ then $x=b=ka_0$ and since $a_0in bar{a}=a+B$ then $a_0=a+b'$ where $b'in B$ then from $b=ka_0$ we have $b=ka+kb'$ hence $b'':=b-kb'=ka$ i.e. $kain B$ but since $bar{a}=a+B$ has order infinite order then $k=0$. Thus $x=0$. Right?
group-theory
group-theory
edited Nov 18 at 3:09
asked Nov 17 at 23:53
ZFR
4,90831337
4,90831337
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1 Answer
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Elements of $A/B$ are cosets. In lemma we suppose that we can write all these cosets as direct sum of $A_{i} / B$, where $A_{i}$ is a subgroup of $A$ and $A_{i}$ has $B$ as a direct summand. This is applied to the situation in theorem as follows:
$,,,,,,,,,,,,,,,,,,,, A / B = bigopluslimits_{i=1}^{n}mathbb{Z}$ (here we keep in mind that copies of $mathbb{Z}$ are numerated)
$,,,,,,$It is enough to find $A_{i}$ as in lemma with $A_{i} = mathbb{Z} oplus B$. But now we can restrict ourselves to the case $pi^{-1}(mathbb{Z}) / B cong mathbb{Z}$, where $pi: A to A / B$ is a canonical projection and $mathbb{Z}$ is the $i$-th copy of $mathbb{Z}$ in $A / B$. This way we can find $A_{i}$ separatly. So this was the reduction to the case $A / B = mathbb{Z}$.
$,,,,,,$About your second question: $ka_{0}$ is a representative of a coset $kB$ in $mathbb{Z} = A / B$, so it is in $B$ only for $k = 0$, meaning that $langle a_0rangle cap B = lbrace 0rbrace$.
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1 Answer
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up vote
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Elements of $A/B$ are cosets. In lemma we suppose that we can write all these cosets as direct sum of $A_{i} / B$, where $A_{i}$ is a subgroup of $A$ and $A_{i}$ has $B$ as a direct summand. This is applied to the situation in theorem as follows:
$,,,,,,,,,,,,,,,,,,,, A / B = bigopluslimits_{i=1}^{n}mathbb{Z}$ (here we keep in mind that copies of $mathbb{Z}$ are numerated)
$,,,,,,$It is enough to find $A_{i}$ as in lemma with $A_{i} = mathbb{Z} oplus B$. But now we can restrict ourselves to the case $pi^{-1}(mathbb{Z}) / B cong mathbb{Z}$, where $pi: A to A / B$ is a canonical projection and $mathbb{Z}$ is the $i$-th copy of $mathbb{Z}$ in $A / B$. This way we can find $A_{i}$ separatly. So this was the reduction to the case $A / B = mathbb{Z}$.
$,,,,,,$About your second question: $ka_{0}$ is a representative of a coset $kB$ in $mathbb{Z} = A / B$, so it is in $B$ only for $k = 0$, meaning that $langle a_0rangle cap B = lbrace 0rbrace$.
add a comment |
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0
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Elements of $A/B$ are cosets. In lemma we suppose that we can write all these cosets as direct sum of $A_{i} / B$, where $A_{i}$ is a subgroup of $A$ and $A_{i}$ has $B$ as a direct summand. This is applied to the situation in theorem as follows:
$,,,,,,,,,,,,,,,,,,,, A / B = bigopluslimits_{i=1}^{n}mathbb{Z}$ (here we keep in mind that copies of $mathbb{Z}$ are numerated)
$,,,,,,$It is enough to find $A_{i}$ as in lemma with $A_{i} = mathbb{Z} oplus B$. But now we can restrict ourselves to the case $pi^{-1}(mathbb{Z}) / B cong mathbb{Z}$, where $pi: A to A / B$ is a canonical projection and $mathbb{Z}$ is the $i$-th copy of $mathbb{Z}$ in $A / B$. This way we can find $A_{i}$ separatly. So this was the reduction to the case $A / B = mathbb{Z}$.
$,,,,,,$About your second question: $ka_{0}$ is a representative of a coset $kB$ in $mathbb{Z} = A / B$, so it is in $B$ only for $k = 0$, meaning that $langle a_0rangle cap B = lbrace 0rbrace$.
add a comment |
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Elements of $A/B$ are cosets. In lemma we suppose that we can write all these cosets as direct sum of $A_{i} / B$, where $A_{i}$ is a subgroup of $A$ and $A_{i}$ has $B$ as a direct summand. This is applied to the situation in theorem as follows:
$,,,,,,,,,,,,,,,,,,,, A / B = bigopluslimits_{i=1}^{n}mathbb{Z}$ (here we keep in mind that copies of $mathbb{Z}$ are numerated)
$,,,,,,$It is enough to find $A_{i}$ as in lemma with $A_{i} = mathbb{Z} oplus B$. But now we can restrict ourselves to the case $pi^{-1}(mathbb{Z}) / B cong mathbb{Z}$, where $pi: A to A / B$ is a canonical projection and $mathbb{Z}$ is the $i$-th copy of $mathbb{Z}$ in $A / B$. This way we can find $A_{i}$ separatly. So this was the reduction to the case $A / B = mathbb{Z}$.
$,,,,,,$About your second question: $ka_{0}$ is a representative of a coset $kB$ in $mathbb{Z} = A / B$, so it is in $B$ only for $k = 0$, meaning that $langle a_0rangle cap B = lbrace 0rbrace$.
Elements of $A/B$ are cosets. In lemma we suppose that we can write all these cosets as direct sum of $A_{i} / B$, where $A_{i}$ is a subgroup of $A$ and $A_{i}$ has $B$ as a direct summand. This is applied to the situation in theorem as follows:
$,,,,,,,,,,,,,,,,,,,, A / B = bigopluslimits_{i=1}^{n}mathbb{Z}$ (here we keep in mind that copies of $mathbb{Z}$ are numerated)
$,,,,,,$It is enough to find $A_{i}$ as in lemma with $A_{i} = mathbb{Z} oplus B$. But now we can restrict ourselves to the case $pi^{-1}(mathbb{Z}) / B cong mathbb{Z}$, where $pi: A to A / B$ is a canonical projection and $mathbb{Z}$ is the $i$-th copy of $mathbb{Z}$ in $A / B$. This way we can find $A_{i}$ separatly. So this was the reduction to the case $A / B = mathbb{Z}$.
$,,,,,,$About your second question: $ka_{0}$ is a representative of a coset $kB$ in $mathbb{Z} = A / B$, so it is in $B$ only for $k = 0$, meaning that $langle a_0rangle cap B = lbrace 0rbrace$.
answered Nov 18 at 12:20
Rybin Dmitry
663
663
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