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Section6.2Lagrange's Theorem

Proof

Corollary 6.12 suggests that groups of prime order \(p\) must somehow look like \({\mathbb Z}_p\).

Remark6.14The converse of Lagrange's Theorem is false

The group \(A_4\) has order 12; however, it can be shown that it does not possess a subgroup of order 6. According to Lagrange's Theorem, subgroups of a group of order 12 can have orders of either 1, 2, 3, 4, or 6. However, we are not guaranteed that subgroups of every possible order exist. To prove that \(A_4\) has no subgroup of order 6, we will assume that it does have such a subgroup \(H\) and show that a contradiction must occur. Since \(A_4\) contains eight 3-cycles, we know that \(H\) must contain a 3-cycle. We will show that if \(H\) contains one 3-cycle, then it must contain more than 6 elements.

In fact, we can say more about when two cycles have the same length.

Proof