If $mathbb{Z}_m times mathbb{Z}_n cong mathbb{Z}_{m'} times mathbb{Z}_{n'}$ with $m|n$ and $m'|n'$, then does...
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Suppose $mathbb{Z}_m times mathbb{Z}_n $ is isomorphic to $mathbb{Z}_{m'} times mathbb{Z}_{n'}$ as groups, where $m$ divides $n$ and $m'$ divides $n'$. Does that mean $m=m'$ and $n=n'$?
group-theory finite-groups cyclic-groups group-isomorphism
edited Dec 18 '18 at 3:06
Shaun
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asked Dec 18 '18 at 2:30
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1707
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closed as off-topic by Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt Dec 18 '18 at 7:31
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$begingroup$
Suppose $mathbb{Z}_m times mathbb{Z}_n $ is isomorphic to $mathbb{Z}_{m'} times mathbb{Z}_{n'}$ as groups, where $m$ divides $n$ and $m'$ divides $n'$. Does that mean $m=m'$ and $n=n'$?
group-theory finite-groups cyclic-groups group-isomorphism
edited Dec 18 '18 at 3:06
Shaun
9,442113684
asked Dec 18 '18 at 2:30
SIONSION
1707
$endgroup$
closed as off-topic by Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt Dec 18 '18 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose $mathbb{Z}_m times mathbb{Z}_n $ is isomorphic to $mathbb{Z}_{m'} times mathbb{Z}_{n'}$ as groups, where $m$ divides $n$ and $m'$ divides $n'$. Does that mean $m=m'$ and $n=n'$?
group-theory finite-groups cyclic-groups group-isomorphism
edited Dec 18 '18 at 3:06
Shaun
9,442113684
asked Dec 18 '18 at 2:30
SIONSION
1707
$endgroup$
Suppose $mathbb{Z}_m times mathbb{Z}_n $ is isomorphic to $mathbb{Z}_{m'} times mathbb{Z}_{n'}$ as groups, where $m$ divides $n$ and $m'$ divides $n'$. Does that mean $m=m'$ and $n=n'$?
group-theory finite-groups cyclic-groups group-isomorphism
group-theory finite-groups cyclic-groups group-isomorphism
edited Dec 18 '18 at 3:06
Shaun
9,442113684
asked Dec 18 '18 at 2:30
SIONSION
1707
edited Dec 18 '18 at 3:06
Shaun
9,442113684
asked Dec 18 '18 at 2:30
SIONSION
1707
edited Dec 18 '18 at 3:06
Shaun
9,442113684
edited Dec 18 '18 at 3:06
Shaun
9,442113684
edited Dec 18 '18 at 3:06
Shaun
9,442113684
9,442113684
asked Dec 18 '18 at 2:30
SIONSION
1707
asked Dec 18 '18 at 2:30
SIONSION
1707
asked Dec 18 '18 at 2:30
SIONSION
1707
1707
closed as off-topic by Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt Dec 18 '18 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt Dec 18 '18 at 7:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Saad, B. Mehta, Trevor Gunn, Derek Holt
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $mathbb Z$-modules
$$
mathbb Z_moplusmathbb Z_ncongmathbb Z_{m'}oplusmathbb Z_{n'}.
$$
Thus for any positive integer $c$ we have
$$
mathbb Z_cotimes_{mathbb Z}(mathbb Z_moplusmathbb Z_n)cong
mathbb Z_cotimes_{mathbb Z}(mathbb Z_{m'}oplusmathbb Z_{n'}).
$$
But
$$
mathbb Z_cotimes_{mathbb Z}mathbb Z_mcongmathbb Z_{gcd(c,m)},
$$
and similarly for the other components. Considering the cardinality of both sides in the above isomorphism,
$$
gcd(c,m)gcd(c,n)=gcd(c,m')gcd(c,n').
$$
Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.
For a more general result see the structure theorem for finitely generated modules
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $mathbb Z$-modules
$$
mathbb Z_moplusmathbb Z_ncongmathbb Z_{m'}oplusmathbb Z_{n'}.
$$
Thus for any positive integer $c$ we have
$$
mathbb Z_cotimes_{mathbb Z}(mathbb Z_moplusmathbb Z_n)cong
mathbb Z_cotimes_{mathbb Z}(mathbb Z_{m'}oplusmathbb Z_{n'}).
$$
But
$$
mathbb Z_cotimes_{mathbb Z}mathbb Z_mcongmathbb Z_{gcd(c,m)},
$$
and similarly for the other components. Considering the cardinality of both sides in the above isomorphism,
$$
gcd(c,m)gcd(c,n)=gcd(c,m')gcd(c,n').
$$
Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.
For a more general result see the structure theorem for finitely generated modules
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
$endgroup$
add a comment |
$begingroup$
Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $mathbb Z$-modules
$$
mathbb Z_moplusmathbb Z_ncongmathbb Z_{m'}oplusmathbb Z_{n'}.
$$
Thus for any positive integer $c$ we have
$$
mathbb Z_cotimes_{mathbb Z}(mathbb Z_moplusmathbb Z_n)cong
mathbb Z_cotimes_{mathbb Z}(mathbb Z_{m'}oplusmathbb Z_{n'}).
$$
But
$$
mathbb Z_cotimes_{mathbb Z}mathbb Z_mcongmathbb Z_{gcd(c,m)},
$$
and similarly for the other components. Considering the cardinality of both sides in the above isomorphism,
$$
gcd(c,m)gcd(c,n)=gcd(c,m')gcd(c,n').
$$
Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.
For a more general result see the structure theorem for finitely generated modules
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
$endgroup$
add a comment |
$begingroup$
Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $mathbb Z$-modules
$$
mathbb Z_moplusmathbb Z_ncongmathbb Z_{m'}oplusmathbb Z_{n'}.
$$
Thus for any positive integer $c$ we have
$$
mathbb Z_cotimes_{mathbb Z}(mathbb Z_moplusmathbb Z_n)cong
mathbb Z_cotimes_{mathbb Z}(mathbb Z_{m'}oplusmathbb Z_{n'}).
$$
But
$$
mathbb Z_cotimes_{mathbb Z}mathbb Z_mcongmathbb Z_{gcd(c,m)},
$$
and similarly for the other components. Considering the cardinality of both sides in the above isomorphism,
$$
gcd(c,m)gcd(c,n)=gcd(c,m')gcd(c,n').
$$
Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.
For a more general result see the structure theorem for finitely generated modules
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
$endgroup$
Yes (assuming $m,n,m',n'$ are positive integers). The group isomorphism produces an isomorphism of $mathbb Z$-modules
$$
mathbb Z_moplusmathbb Z_ncongmathbb Z_{m'}oplusmathbb Z_{n'}.
$$
Thus for any positive integer $c$ we have
$$
mathbb Z_cotimes_{mathbb Z}(mathbb Z_moplusmathbb Z_n)cong
mathbb Z_cotimes_{mathbb Z}(mathbb Z_{m'}oplusmathbb Z_{n'}).
$$
But
$$
mathbb Z_cotimes_{mathbb Z}mathbb Z_mcongmathbb Z_{gcd(c,m)},
$$
and similarly for the other components. Considering the cardinality of both sides in the above isomorphism,
$$
gcd(c,m)gcd(c,n)=gcd(c,m')gcd(c,n').
$$
Recall that $c$ was arbitrary. Setting $c=m$ gives $m|m'$ and $c=m'$ gives $m'|m$, so $m=m'$. Now setting $c=n$ gives $n|n'$ and $c=n'$ gives $n'|n$, so $n=n'$.
For a more general result see the structure theorem for finitely generated modules
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
answered Dec 18 '18 at 3:46
stewbasicstewbasic
5,7531926
5,7531926
add a comment |
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