Question

prove that a group fode 105 contain
a subegroup forder35​

Answer 1

Step-by-step explanation:

The group G has order |G|=105=3⋅5⋅7. If np is the number of Sylow-p subgroups, then n3∈{1,7}, n5∈{1,21}, and n7∈{1,15}.

By Cauchy's theorem, G has subgroups K1, K2 of order 5 and 7 respectively. If neither of these are normal, then n5=21 and n7=15. But this gives us 175 elements, which is impossible. Thus either K1 or K2 is normal, so K1K2 is a subgroup of order |K||H||H∩K|=35.

Now write H=K1K2. Then G acts transitively on the cosets G/H, so we have a map G→S3. If this map is trivial, then H is normal in G. If not, then the image has order 3 (since |G| and 2 are coprime), so the kernel is a normal subgroup of order 35. If x is in the kernel, then xH=H, so x∈H. That is, the kernel is exactly H, so again H is normal in G.

From Sylow theory, G acts transitively on all Sylow-p subgroups via conjugation. But any conjugate of K1 lies in H since H is normal. Since H is necessarily cyclic, K1 is normal in H, so H contains 1 Sylow-5 subgroup. This means G also contains only 1, and similarly for the Sylow-7 group.

Finally, note that it is not necessarily true that n3=1, because there is a nonabelian group of order 105. You can construct this from any nontrivial automorphism C3→Aut(C5×C7)≅(Z/35)×. One exists since |(Z/35)×|=(5−1)(7−1)=24=3⋅23, so we can send a generator of C3 to the element of order 3.

Answer 2

Answer:

You need to use Sylow theorem

properly. Since 105=3⋅5⋅7, the

number 1+7k of 7-Sylow subgroups

is 1 or 15. If there are 15 such

subgroups then we get 15⋅6=90

elements of order 7. This makes it impossible for the number 1+5k

of 5-Sylow subgroups to be 6 or

more, since that would require

at least 6⋅4=24 elements of order 5, making the group too big. Thus

there is a unique subgroup G5 of

order 5, and it is normal. For any

7-Sylow subgroup G7 we thus

have that G5G7 is a subgroup

of order 35, and its order is 35

because G5∩G7=1. (In general, the intersection of a p-Sylow subgroup

and a q-Sylow subgroup for p≠q

must be trivial.) On the other hand,

if there a unique subgroup G7 of

order 7, then it is normal. For any

5-Sylow subgroup G5 we thus

have that G5G7 is a subgroup of

order 35 for the same reasons

as in the previous paragraph but

with the roles of 5 and 7 exchanged.