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How to find $(mathbb Z/12times mathbb Z/12)/$?
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Consider $G=mathbb Z/12times mathbb Z/12$ and its subgroup $H$ generated by $(a^4,a^6)$, where $a$ is a generator of $mathbb Z/12$. How do I find $G/H$ as a product of cyclic groups of prime power orders?
I know how to identify similar quotients in the case $mathbb Ztimes mathbb Z$ via reducing a matrix to the Smith normal form (see this answer for instance), but I don't know whether such technique carries over to this case.
abstract-algebra group-theory finite-groups abelian-groups quotient-group
asked Jul 19 at 5:59
user437309
556212
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up vote
2
down vote
favorite
Consider $G=mathbb Z/12times mathbb Z/12$ and its subgroup $H$ generated by $(a^4,a^6)$, where $a$ is a generator of $mathbb Z/12$. How do I find $G/H$ as a product of cyclic groups of prime power orders?
I know how to identify similar quotients in the case $mathbb Ztimes mathbb Z$ via reducing a matrix to the Smith normal form (see this answer for instance), but I don't know whether such technique carries over to this case.
abstract-algebra group-theory finite-groups abelian-groups quotient-group
asked Jul 19 at 5:59
user437309
556212
3
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider $G=mathbb Z/12times mathbb Z/12$ and its subgroup $H$ generated by $(a^4,a^6)$, where $a$ is a generator of $mathbb Z/12$. How do I find $G/H$ as a product of cyclic groups of prime power orders?
I know how to identify similar quotients in the case $mathbb Ztimes mathbb Z$ via reducing a matrix to the Smith normal form (see this answer for instance), but I don't know whether such technique carries over to this case.
abstract-algebra group-theory finite-groups abelian-groups quotient-group
asked Jul 19 at 5:59
user437309
556212
Consider $G=mathbb Z/12times mathbb Z/12$ and its subgroup $H$ generated by $(a^4,a^6)$, where $a$ is a generator of $mathbb Z/12$. How do I find $G/H$ as a product of cyclic groups of prime power orders?
I know how to identify similar quotients in the case $mathbb Ztimes mathbb Z$ via reducing a matrix to the Smith normal form (see this answer for instance), but I don't know whether such technique carries over to this case.
abstract-algebra group-theory finite-groups abelian-groups quotient-group
asked Jul 19 at 5:59
user437309
556212
asked Jul 19 at 5:59
user437309
556212
asked Jul 19 at 5:59
user437309
556212
asked Jul 19 at 5:59
user437309
556212
556212
3
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46
add a comment |Â
3
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46
3
3
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46
add a comment |Â
1 Answer
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up vote
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Note that $H=langle(a^4,0),(0,a^6)rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$Bbb Z/12timesBbb Z/12/langle(a^4,0),(0,a^6)rangle\cong big(Bbb Z/12timesBbb Z12/langle(a^4,0)ranglebig)/langle(0,a^6)rangle\congBbb Z/4timesBbb Z/6$$
answered Jul 19 at 6:09
Arthur
98.8k793175
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that $H=langle(a^4,0),(0,a^6)rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$Bbb Z/12timesBbb Z/12/langle(a^4,0),(0,a^6)rangle\cong big(Bbb Z/12timesBbb Z12/langle(a^4,0)ranglebig)/langle(0,a^6)rangle\congBbb Z/4timesBbb Z/6$$
answered Jul 19 at 6:09
Arthur
98.8k793175
add a comment |Â
up vote
3
down vote
accepted
Note that $H=langle(a^4,0),(0,a^6)rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$Bbb Z/12timesBbb Z/12/langle(a^4,0),(0,a^6)rangle\cong big(Bbb Z/12timesBbb Z12/langle(a^4,0)ranglebig)/langle(0,a^6)rangle\congBbb Z/4timesBbb Z/6$$
answered Jul 19 at 6:09
Arthur
98.8k793175
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that $H=langle(a^4,0),(0,a^6)rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$Bbb Z/12timesBbb Z/12/langle(a^4,0),(0,a^6)rangle\cong big(Bbb Z/12timesBbb Z12/langle(a^4,0)ranglebig)/langle(0,a^6)rangle\congBbb Z/4timesBbb Z/6$$
answered Jul 19 at 6:09
Arthur
98.8k793175
Note that $H=langle(a^4,0),(0,a^6)rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$Bbb Z/12timesBbb Z/12/langle(a^4,0),(0,a^6)rangle\cong big(Bbb Z/12timesBbb Z12/langle(a^4,0)ranglebig)/langle(0,a^6)rangle\congBbb Z/4timesBbb Z/6$$
answered Jul 19 at 6:09
Arthur
98.8k793175
edited Jul 19 at 6:24
answered Jul 19 at 6:09
Arthur
98.8k793175
answered Jul 19 at 6:09
Arthur
98.8k793175
answered Jul 19 at 6:09
Arthur
98.8k793175
98.8k793175
add a comment |Â
add a comment |Â
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3
You can use the Smith algorithm once you recognise you can step back and then consider $mathbbZtimesmathbbZ$ factored by $langle (12,0), (0,12), (4,6)rangle$.
– ancientmathematician
Jul 19 at 6:46