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Then P d" PQ and Q d" PQ. But then P is an element of S and Q is an element of S. But recall Q is an element of T. This contradicts that S )" T is the empty set, and so PQ is not in S. Similarly, PQ is not an element of T. Thus, PQ is not a proper subgroup of G, and PQ = G. By Lemma 2.4, G is isomorphic to PQ. Lastly, we claim p `" q. Suppose p = q. Let P = <x> and Q = <y>. Since x and y have order p, the subgroups <(x,e)>, <(e,y)>, and <(x,y)> of PQ have order p. But G only has two subgroups of prime order, namely P and Q, and G is isomorphic to PQ. This is a contradiction, and so p `" q. Therefore, G is isomorphic to EMBED Equation.3 , which is isomorphic to EMBED Equation.3 , completing the proof. The reader should observe that we may adjust Definition 1.4 to define a group whose subgroup lattice is formed by n chains for any positive integer n. For example, for any prime p, EMBED Equation.3 would be a group whose subgroup lattice is formed by p+1 chains. The reason for this is that by Lagranges Theorem, all nontrivial subgroups of Zp x Zp have order p. There are p+1 such subgroups of Zp x Zp. If Zp = <{0,1,2,, p-1}, + mod p>, then the p+1 subgroups are <(1,0)> and {<(x,1)> |RZH$7$8$H$a$gda-$a$gd};`gdL~gddD`gd2`gddD&LZް"$JLNPR1:?luȻ香ݕ}umjh}LUh}Lh.i6h}Lh}L6h}Lh.ih$ch hdDjhkh.iEHU)j H h.iCJPJUVaJnH tH jnhkh.iEHU)j H h.iCJPJUVaJnH tH jhkUhhkh[}]h h h|dhP%ٲ:DESVXYZ[\]y~ӳڳ۳YZcͫͤͫͫͤͫ͠~y h)5hL~h$cUhL~hghLh@ hnH*hnhrhXxhXx hXxH* h 616h 61h 616h$ch 61h2h}L5hXxh}Ljh}LUjhkh}LEHU)jMH h}LCJPJUVaJnH tH / x ranges over {0,1,,p-1}}. Example 3.5 demonstrates what happens in the case that p=2. Example 3.5. The subgroup lattice of EMBED Equation.3 is formed by three chains. SHAPE \* MERGEFORMAT REFERENCES 1. David S. Dummit and Richard M. Foote. Abstract Algebra, Prentice Hall, 1991. 2. Joseph A. Gallian. Contemporary Abstract Algebra, Houghton Mifflin Company, 2002. 3. Roland Schmidt. 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