most recent 30 from http://mathoverflow.net 2013-05-24T01:03:49Z http://mathoverflow.net/feeds/user/15767 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67743/maximal-abelian-subgroups-of-p-groups unknown (google) 2011-06-14T08:18:45Z 2011-06-14T09:23:08Z

<p>A non-abelian group of order $p^n$ ($n\geq 4$) always has normal abelian group of order $p^3$, and this theorem is useful in enumeration/ classification of groups of order $p^4$. So, abelian normal subgroups of $p$ groups are useful in the classification problem.</p> <p>Alperin, in his paper on "Large Abelian Subgroups of $p$ groups" stated a result of Burnside namely </p> <p>"<em>a group of order $p^n$ has normal abelian subgroups of order $p^m$ with $n\leq m(m-1)/2$</em>".</p> <p>Question: For (non-abelian) group $G$ of order $2^5$, by result of Burnside, there will be normal abelian subgroups of order $p^m$ with $5\leq m(m-1)/2$, which means $m\geq 4$. So conclusion is $G$ always has normal abelian subgroup of order $2^4$. But if we check the list of groups of order $2^5$, then there are some non-abelian groups where maximaum order of abelian (normal) subgroup is $2^3$. </p> <p>Can one explain, what is going wrong here? (I am confused with this theorem.)</p> <p>Does all maximal abelian subgroups of a non-abelian finite $p$ group have same order?</p> <p>Also, please, suggest some reference for some results on maximal abelian subgroups of $p$ groups?</p>

http://mathoverflow.net/questions/67743/maximal-abelian-subgroups-of-p-groups 2011-06-14T08:38:09Z 2011-06-14T08:38:09Z

@Someone: I went through some papers of Alperin and Burnside, but still I am not satisfied. I didn't get enough material. If someone gives some direction for these questions, then its fine.