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Fun Exercise: Let \(G\) be a group of order \(p^2\), where \(p\) is prime. Show that \(G\) has a subgroup of order \(p\).
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\(p^2\) must divide the order of any element in \(G\). So the order of elements in \(G\) are restricted to \(1, p, p^2\). Consider a non identity element \(a\) of \(G\); the order can be \(p\) or \(p^2\). If the order of \(a\) is \(p\), then we are done. otherwise the order of \(a^p\) will be \(p\).
You just breeze through these, don't you? :-P Here's your medal.
Haha.. to be honest idk much of abstract algebra... The questions you have been asking have some overlap with NT. My textbook covers them under primitive roots and quadratic reciprocity law...
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