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But recall g[Q] = Q. Since g is an injection, g[P] = P. So for all g in G, g[P] = P and P is normal by Lemma 3.3. By a similar argument, Q is normal. Now that P and Q are normal, by Lemma 3.4, PQ d" G. We wish to show PQ = G. Suppose PQ is a proper subgroup of G. Then PQ is an element of S or PQ is an element of T. Suppose PQ is an element of S. 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> where x and y are elements of G. Then < (x,y) > has prime order. But P and Q are the only subgroups of prime order. This is a contradiction, and so p `"q. Therefore, G is isomorphic to EMBED Equation.3 , which is isomorphic toN$:<TVΠР &<RVX\bfҡ>46DF֣أFLNvƤJXt6n`~ȧ۹jhkUhkhd9h h 5\h}[h h&;hm@hhwhI7h h mHsHh h hkVU@ȧʧ̧ΧЧ/012qzƬȬBKʳբΙydWyR h)5jhkh}LEHU)jMH h}LCJPJUVaJnH tH jh}LUh}Lh.i6h}Lh}L6h}Lh.i h/5h hdDjhkh.iEHU)j H h.iCJPJUVaJnH tH Uhkh h jhkUjhkh.iEHU)j H h.iCJPJUVaJnH tH 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. Example 3.5. The subgroup lattice of EMBED Equation.3 is formed by three chains. SHAPE \* MERGEFORMAT REFERENCES [DF] David S. Dummit and Richard M. Foote. Abstract Algebra, Prentice Hall, 1991. [G] Joseph A. Gallian. Contemporary Abstract Algebra, Houghton Mifflin Company, 2002. [S] Roland Schmidt. 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