Question

The following exercise shows that the converse to Lagrange's theorem is false, i.e. even if d ||G|, there need not be a subgroup of G with order d. (a) Let n > 4 and consider the alternating group An. Suppose that NC An is a normal subgroup and that there is a 3-cycle (abc) E N. Prove that N = An. Hint: it is enough to show that N contains all 3-cycles. What is the conjugate of (abc) by the double-transposition (ab)(cd)? (b) Use (2b) to prove that A4 contains no subgroup of order 6.

This is 2(b):

Answer 1

026 Suppose that [G! H] = 2 . Prove that His a normal шь оиф о сп. het H be a subgroup of inden 2 in a group o. het ett be the identity element of G. To prove that H is a normal subgroup, we want to show that for any get, gH = Hg. If g EH, then this is true so we assume that g & H. Note that left losets partition a unto two disjoint sets since the index is 2. since 94 H these are ett and g H. sime early right cosets partition or into two disjoint sets. These disjoint right cosets are He and Hg. Because of these partitions, we have as sets gu = G-eH = G H = G-He = Mg tas stanned fritts a normal subgroup in G. CamScanner