Show that there are two abelian groups of order 108 that have exactly one subgroup of order 3.(adsbygoogle = window.adsbygoogle || []).push({});

108 = 2^ 2 X 3 ^ 3

Using the fundamental theorem of finite abelian groups, we have

Possible abelian groups of order 108 can be : Z108, Z4 + Z27, Z2+Z2+Z27, Z4+Z9+Z3, Z2+Z2+Z9+Z3, Z4+Z3+Z3+Z3, Z2+Z2+Z3+Z3+Z3

It seems to me that all three Z108, Z4+Z27, Z2+Z2+Z27 have exactly one subgroup of order 3. Please suggest where am I going wrong ?

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# Fundamental Theorem of Abelian Groups

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