Yahoo Answers: Answers and Comments for Abstract Algebra  Cyclic Normal Subgroups? [Mathematics]
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From Anonymous
enUS
Mon, 31 Oct 2011 19:17:35 +0000
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Yahoo Answers: Answers and Comments for Abstract Algebra  Cyclic Normal Subgroups? [Mathematics]
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From David: the first thing to do is factor 385 = (5)(7)(1...
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https://answers.yahoo.com/question/index?qid=20111031191734AAbaqr3
Tue, 01 Nov 2011 03:56:27 +0000
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 coprime), 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 11sylow 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 11sylow 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 5sylow 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.