Best Answer:
the first thing to do is factor 385 = (5)(7)(11).

by cauchy's theorem, we know G has an element of order 7 and an element of order 11, say x and y. thus H = <x> and K = <y> are subgroups of order 7, and 11, respectively.

since H∩K = {e} (since 7 and 11 are co-prime), HK (if it is a group) has order 77.

if we can show that either H or K is normal, then HK will indeed be a subgroup.

now K is an 11-sylow subgroup, and the number of such subgroups = 1 (mod 11).

the number of such subgroups also divides 385, which has divisors:

1,5,7,11,35,55,77,385. of these, only 1 is congruent to 1 mod 11, so there is only

1 11-sylow subgroup, which must therefore be normal.

thus HK is indeed a subgroup, which is of index 5.

is H normal as well? again, H is a 5-sylow subgroup, so the number of such

subgroups is congruent to 1 (mod 5) and divides 385.

again, we see 1 is the only possibility, so H is normal as well.

thus, for any g in G, and hk in HK,

g(hk)g^-1 = (ghg^-1)(gkg^-1) = h'k', so HK is normal in G.

now HK is the internal direct product of two cyclic groups,

so it's isomorphic to Z7xZ11, which by the chinese remainder theorem

is isomorphic to Z77, so HK is a cyclic normal subgroup of index 5.

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