- #1
Jösus
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Hello
Lately, I have been studying some group theory. On my own, I should add, so I don't really have any professor (or other knowledgeable person for that matter) to ask when a problem arises; which is why I am here.
I had set out to find all small groups (up to order 30 or something), up to isomorphism. Up until the unique group of order 15, no problem arises. However, at order 16 I have gotten stuck. Naturally, the abelian ones poses no real challange, but as only one prime is involved, Sylow Theory seems to be of little help, and so seems any other method I have tried. I would greatly appreciate some small hints on how to proceed, but I would like to do the rest myself. I am in the process of dropping my goal of finding the groups of said orders, but this one has been bugging me. So has the next one, but there I at least think I got off to a decent start.
The next problematic order is 21 (and a whole bunch of others with similar factorisation into primes). I try to approach it as follows: Sylow Theory tells us that if G is a finite group, p divides the order of G and p is a prime, then |Syl_{p}(G)| \equiv 1 (mod \: p) and |Syl_{p}(G)| divides |G|. From this, and the fact that 3 is congruent to 3 modulo 7, we draw the conclusion that any group of order 21 contains a normal subgroup of order 7. This subgroup is, as it is of prime order, a cyclic one. Moreover, the intersection of this normal subgroup and any Sylow 3-subgroup is simply the singleton \{1_{G}\}. Also, if we by S and T denote the groups of order 3 and 7, respectively, we have ST = \{st | s \in S, t \in T\} = G. Thus G is the semidirect product T \rtimes S. Now, as the involved groups are cyclic, this seems to be an easy match from here on. Here I guess I should give a presentation for the group, and it would be something like G = \langle a, b | a^7 = b^3 = 1, b^{-1}ab = a^k \rangle where k is one of the integers 1, ... ,6. k = 1 yields the abelian group (the cyclic) of order 21. Another one would be the group with presentation \langle a, b | a^7 = b^3 = 1, b^{-1}ab = a^2 \rangle.
Here I am stuck. There seems to be too many possibilities for it to be reasonable to just play around with the relations and try to look for equalities between the different products. I know, I should really read more about presentations, but I don't have any good litterature for the moment. If anyone would feel like helping out, tell me if I made some terrible mistakes and faulty assumptions or anything else that could help me get a grasp on what I am doing, I would be very thankful.
Thanks in advance!
Lately, I have been studying some group theory. On my own, I should add, so I don't really have any professor (or other knowledgeable person for that matter) to ask when a problem arises; which is why I am here.
I had set out to find all small groups (up to order 30 or something), up to isomorphism. Up until the unique group of order 15, no problem arises. However, at order 16 I have gotten stuck. Naturally, the abelian ones poses no real challange, but as only one prime is involved, Sylow Theory seems to be of little help, and so seems any other method I have tried. I would greatly appreciate some small hints on how to proceed, but I would like to do the rest myself. I am in the process of dropping my goal of finding the groups of said orders, but this one has been bugging me. So has the next one, but there I at least think I got off to a decent start.
The next problematic order is 21 (and a whole bunch of others with similar factorisation into primes). I try to approach it as follows: Sylow Theory tells us that if G is a finite group, p divides the order of G and p is a prime, then |Syl_{p}(G)| \equiv 1 (mod \: p) and |Syl_{p}(G)| divides |G|. From this, and the fact that 3 is congruent to 3 modulo 7, we draw the conclusion that any group of order 21 contains a normal subgroup of order 7. This subgroup is, as it is of prime order, a cyclic one. Moreover, the intersection of this normal subgroup and any Sylow 3-subgroup is simply the singleton \{1_{G}\}. Also, if we by S and T denote the groups of order 3 and 7, respectively, we have ST = \{st | s \in S, t \in T\} = G. Thus G is the semidirect product T \rtimes S. Now, as the involved groups are cyclic, this seems to be an easy match from here on. Here I guess I should give a presentation for the group, and it would be something like G = \langle a, b | a^7 = b^3 = 1, b^{-1}ab = a^k \rangle where k is one of the integers 1, ... ,6. k = 1 yields the abelian group (the cyclic) of order 21. Another one would be the group with presentation \langle a, b | a^7 = b^3 = 1, b^{-1}ab = a^2 \rangle.
Here I am stuck. There seems to be too many possibilities for it to be reasonable to just play around with the relations and try to look for equalities between the different products. I know, I should really read more about presentations, but I don't have any good litterature for the moment. If anyone would feel like helping out, tell me if I made some terrible mistakes and faulty assumptions or anything else that could help me get a grasp on what I am doing, I would be very thankful.
Thanks in advance!
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