Classify all groups of order $3times 5times 67$
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I'm working on classifying all groups of order $3*5*67= 1005$ and I've made some progress but I'm not fully there yet.
First, I've found that the sylow-67= $S_67$ subgroup is normal by using sylows theorems. I'd like to show that there exists a subgroup of order 15. I noticed that $G/S_67$ is of order 15 so it must be abelian(we can use sylows theorems to show this). I'm wondering if that is helpful somehow.
Once we establish that there is such a group of order 15, then we can simply say that $G = mathbbZ/67 rtimes mathbbZ/15$, and then see that there are only two possible groups. The abelian group, and $$<a,b|a^67 = 1, b^15 = 1, bab^-1 = a^22>$$
abstract-algebra group-theory
asked Aug 12 at 4:03
iYOA
60549
add a comment |Â
up vote
1
down vote
favorite
I'm working on classifying all groups of order $3*5*67= 1005$ and I've made some progress but I'm not fully there yet.
First, I've found that the sylow-67= $S_67$ subgroup is normal by using sylows theorems. I'd like to show that there exists a subgroup of order 15. I noticed that $G/S_67$ is of order 15 so it must be abelian(we can use sylows theorems to show this). I'm wondering if that is helpful somehow.
Once we establish that there is such a group of order 15, then we can simply say that $G = mathbbZ/67 rtimes mathbbZ/15$, and then see that there are only two possible groups. The abelian group, and $$<a,b|a^67 = 1, b^15 = 1, bab^-1 = a^22>$$
abstract-algebra group-theory
asked Aug 12 at 4:03
iYOA
60549
2
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on classifying all groups of order $3*5*67= 1005$ and I've made some progress but I'm not fully there yet.
First, I've found that the sylow-67= $S_67$ subgroup is normal by using sylows theorems. I'd like to show that there exists a subgroup of order 15. I noticed that $G/S_67$ is of order 15 so it must be abelian(we can use sylows theorems to show this). I'm wondering if that is helpful somehow.
Once we establish that there is such a group of order 15, then we can simply say that $G = mathbbZ/67 rtimes mathbbZ/15$, and then see that there are only two possible groups. The abelian group, and $$<a,b|a^67 = 1, b^15 = 1, bab^-1 = a^22>$$
abstract-algebra group-theory
asked Aug 12 at 4:03
iYOA
60549
I'm working on classifying all groups of order $3*5*67= 1005$ and I've made some progress but I'm not fully there yet.
First, I've found that the sylow-67= $S_67$ subgroup is normal by using sylows theorems. I'd like to show that there exists a subgroup of order 15. I noticed that $G/S_67$ is of order 15 so it must be abelian(we can use sylows theorems to show this). I'm wondering if that is helpful somehow.
Once we establish that there is such a group of order 15, then we can simply say that $G = mathbbZ/67 rtimes mathbbZ/15$, and then see that there are only two possible groups. The abelian group, and $$<a,b|a^67 = 1, b^15 = 1, bab^-1 = a^22>$$
abstract-algebra group-theory
asked Aug 12 at 4:03
iYOA
60549
asked Aug 12 at 4:03
iYOA
60549
asked Aug 12 at 4:03
iYOA
60549
asked Aug 12 at 4:03
iYOA
60549
60549
2
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18
add a comment |Â
2
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18
2
2
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18
add a comment |Â
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2
I think it might make sense to consider the subgroup generated by an element of order $5$ and an element of order $67$; this is cyclic [can you see why?]. Then you can write your group as $mathbbZ_335rtimesmathbbZ_3$.
– Steve D
Aug 12 at 4:13
oh wow! that was slick. I was barking up the wrong tree
– iYOA
Aug 12 at 4:15
But yes there is an element of order $15$, because the subgroup of order $335$ is cyclic, hence has a normal Sylow 5-subgroup. Sylow's theorems then show there is only one Sylow 5-subgroup in your group.
– Steve D
Aug 12 at 4:15
Good point. Any ideas on how to show there is an element of order 15 without using the cyclic group of order 335? I'm wondering if there was any way to salvage the previous argument.
– iYOA
Aug 12 at 4:18