Finite nonabelian groups of odd order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:33:20Z http://mathoverflow.net/feeds/question/31553 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31553/finite-nonabelian-groups-of-odd-order Finite nonabelian groups of odd order falagar 2010-07-12T14:06:10Z 2010-07-14T04:41:09Z <p>For every even $n$ there exists nonabelian group. As example of such group we can take dihedral group.</p> <p>The question is about odd $n$. For some of them there are no nonabelian groups of order $n$ (for example, if $n$ is prime then the group of order $n$ is cyclic and hence abelian).</p> <p>For what odd $n$ are there known examples of nonabelian finite groups of order $n$?</p> http://mathoverflow.net/questions/31553/finite-nonabelian-groups-of-odd-order/31558#31558 Answer by Robin Chapman for Finite nonabelian groups of odd order Robin Chapman 2010-07-12T14:29:05Z 2010-07-13T00:26:06Z <p>It's well-known that for a natural number $n$ with prime factorization $n=\prod_i p_i^{r_i}$, all groups of order $n$ are abelian if and only if all $r_i\le 2$ and $\gcd(n,\Phi(n))=1$ where $\Phi(n)=\prod_i (p_i^{r_i}-1)$. (See <a href="http://groups.google.co.uk/group/sci.math/msg/215efc43ebb659c5?hl=en" rel="nofollow">http://groups.google.co.uk/group/sci.math/msg/215efc43ebb659c5?hl=en</a>)</p> <p>For other $n$ there are non-abelian groups. If some $r_i\ge3$ then we can take a direct product of a non-abelian group of order $p_i^3$ and a cyclic group. There are always non-abelian groups of order $p^3$; when $p=2$ take the quaternion group, and when $p$ is odd the group of upper triangular matrices with unit diagonal over $\mathbb{F}_p$.</p> <p>Otherwise $G$ will have a factor $pq$ with $p\mid(q-1)$ or $pq^2$ with $p\mid(q^2-1)$. In the first case the group of all maps $x\mapsto ax+b$ for $a$, $b$, $x\in\mathbb{F}_q$ and $a\ne 0$ has a non-abelian subgroup of order $pq$. In the second case replace $\mathbb{F}_q$ by <code>$\mathbb{F}_{q^2}$</code> and then get a non-abelian group of order $pq^2$. In both cases multiply by a cyclic group to get an order $n$ non-abelian group.</p> http://mathoverflow.net/questions/31553/finite-nonabelian-groups-of-odd-order/31793#31793 Answer by Amitesh Datta for Finite nonabelian groups of odd order Amitesh Datta 2010-07-14T04:41:09Z 2010-07-14T04:41:09Z <p>You might be interested in the result that if <em>n</em> is odd, |<em>G</em>| = <em>n</em> for a finite group <em>G</em>, and if every subgroup of <em>G</em> is normal, then <em>G</em> is abelian. (This does not hold if the hypothesis that <em>n</em> is odd is ommitted as the quaternion group of order 8 demonstrates.) </p> <p>A group whose every subgroup is normal is called a <em>Dedekind group</em>. A non-abelian Dedekind group is called a <em>Hamiltonian group</em>. With this terminology the result simply states that a Dedekind group of odd order is abelian. </p> <p>The proof is not immediately obvious. It relies on a classification result that states that every Hamiltonian group is a direct product of the quaternion group of order 8, an elemetary abelian 2-group, and a periodic abelian group of odd order. Once this classification result is established, however, the result can be seen easily.</p>