Maximal Abelian Subgroups of p-groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:33:42Z http://mathoverflow.net/feeds/question/67743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67743/maximal-abelian-subgroups-of-p-groups 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/67745#67745 Answer by Derek Holt for Maximal Abelian Subgroups of p-groups Derek Holt 2011-06-14T09:23:08Z 2011-06-14T09:23:08Z <p>As I explained in my answer on maths.stackexchange, what Alperin wrote is clearly wrong. He has misquoted what Burnside proved, which was that a group of order $p^n$ with centre of order $p^c$ contains a normal abelian subgroup of order $p^m$ for some $m$ with $n≤m+(m−c)(m+c−1)/2$. Burnside cites a related result of Miller that there is a normal abelian subgroup of order $p^m$, for any $m$ with $n>m(m−1)/2$. What is that you are still confused about?</p> <p>The answer to your second question is no. For example a dihedral group of order 16 has a maximal cyclic subgroup of order 8, but it also has subgroups of order 4 isomorphic to $C_2^2$, which are maximal subject to being abelian.</p>