Finite groups with elements of the same order - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:14:22Z http://mathoverflow.net/feeds/question/39848 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order Finite groups with elements of the same order Denis Serre 2010-09-24T11:51:29Z 2010-09-24T12:01:18Z <p>Given a finite group $G$, let ${(1,1),(m_1,n_1),\ldots,(m_r,n_r)}$ be the list of pairs $(m,n)$ in which $m$ is the order of some element, and $n$ is the number of elements with this order. The order of $G$ is thus $1+n_1+\cdots+n_r$, and the pair $(1,1)$ accounts for the neutral element.</p> <p>Let $G,G'$ be two finite groups, with the same list. Is it true or not (I bet <em>not</em>) that $G$ and $G'$ are isomorphic ? If not, please provide a counter-exemple. </p> http://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order/39850#39850 Answer by Bill Thurston for Finite groups with elements of the same order Bill Thurston 2010-09-24T12:01:05Z 2010-09-24T12:01:05Z <p>There are easy examples that are $p$-groups. For instance, the mod 3 Heisenberg group is the nilpotent group with presentation <code>$\left &lt; a,b,c \;\bigg |\, [a,b] = c, [a,c] = [b,c] = a^3 = b^3 = c^3 = 1 \right &gt;$</code> has order 27, and all but the trivial element of order 3. This has the same order portrait as $C_3^3$ where $C_3 = \mathbb Z / 3\mathbb Z$ is the cyclic group of order 3.</p> http://mathoverflow.net/questions/39848/finite-groups-with-elements-of-the-same-order/39851#39851 Answer by Colin Reid for Finite groups with elements of the same order Colin Reid 2010-09-24T12:01:18Z 2010-09-24T12:01:18Z <p>See this question and the first answer:</p> <p><a href="http://mathoverflow.net/questions/31249/order-information-enough-to-guarantee-1-isomorphism" rel="nofollow">http://mathoverflow.net/questions/31249/order-information-enough-to-guarantee-1-isomorphism</a></p> <p>A fortiori, any counterexample given to that question will work for your question as well.</p>