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lola1990
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Homework Statement
I am having problems with parts c and dHomework Equations
The Attempt at a Solution
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lola1990 said:But, why can't it be 231?
lola1990 said:Oh wait, I messed up the first part. The two possibilities for the kernel are 1 and 7, and if it is 1 then |C/kernel|=7. Since the number of permutations of P is 9!, I'm unclear on why this leads to contradiction. Is there way to know |Aut(P)|<9! ?
lola1990 said:Oh, I do have a theorem that the automorphism group of the cyclic group of order n is isomorphic to (Z/nZ)^x. So, since the co-prime elements of Z/9Z are 1,2,4,5,7,8 the automorphism group of Z9 has order 6.
Likewise, the automorphism group pf Z3 has order 3. Does this mean that the automorphism group of Z3 x Z3 has order 9?
lola1990 said:Just want to check that 0 is not an element of Z/9Z, right? Because then the order would be 7 and the proof would go out the window.
What do you mean by Z3 linear? I guess if f is an automorphism, then f(x+y)=f(x)f(y). Not quite sure if this is what you mean...
lola1990 said:ooh... so ax=x+x+..+x (a times), s0 f(x,y)+f(x,y)...+f(x,y) (a times)=f((x,y)+..+(x,y) (a times))=f(a(x,y)). Is that right? So then, af(x,y)+bf(z,w)=f(a(x,y))+f(b(z,w))=f(a(x,y)+b(z,w)). Thus, if f is an automorphism, it is a linear map in Z3xZ3. Do I also need to show that a linear map in Z3x Z3 is an automorphism of Z3xZ3?
lola1990 said:No, I do see why a bijective linear map is an automorphism. I know that the automorphism induces a linear map from Z3 x Z3 to Z3 x Z3, but do I know that it is bijective? Also, does a 2x2 matrix define a bijective linear map or just a linear map?
lola1990 said:Clearly, there are 9 possibilities for choosing the first vector (x,y).
A Sylow p-subgroup is a subgroup of a finite group whose order is a power of a prime number p.
The number of Sylow p-subgroups in a group of order 693 can be found using Sylow's third theorem, which states that the number of Sylow p-subgroups is congruent to 1 modulo p. In this case, since 693 = 3 x 7 x 11, there can be either 1 or 13 Sylow 3-subgroups, either 1 or 99 Sylow 7-subgroups, and either 1 or 77 Sylow 11-subgroups.
Sylow p-subgroups provide information about the prime factors of the order of a group and can help in determining its normal subgroups. In the case of a group of order 693, the Sylow 3-subgroups, 7-subgroups, and 11-subgroups can provide insight into the structure of the group and its possible subgroups.
No, according to Sylow's third theorem, the number of Sylow p-subgroups is congruent to 1 modulo p. Since 693 = 3 x 7 x 11, the number of Sylow 7-subgroups must be either 1 or 99. Therefore, a group of order 693 cannot have more than one Sylow 7-subgroup.
No, Sylow p-subgroups are not unique in a group of order 693. As mentioned before, there can be either 1 or 13 Sylow 3-subgroups, either 1 or 99 Sylow 7-subgroups, and either 1 or 77 Sylow 11-subgroups. Therefore, there can be multiple Sylow p-subgroups for each prime factor of the order of the group.