faithful unipotent representations of (finite) $p$-groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:57:23Z http://mathoverflow.net/feeds/question/55685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups faithful unipotent representations of (finite) $p$-groups Igor Rivin 2011-02-17T02:15:45Z 2012-09-06T17:38:56Z <p>The title pretty much summarizes the question: does every $p$-group have a <em>faithful</em> unipotent representation (with coefficients in $\mathbb{F}_p$ or some finite extension thereof)? </p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55686#55686 Answer by Mark Sapir for faithful unipotent representations of (finite) $p$-groups Mark Sapir 2011-02-17T03:12:11Z 2011-02-18T02:17:07Z <p>Yes, of course, it is how Bogopolsky proves Sylow theorems in his book (Bogopolski, Oleg, Introduction to group theory. Translated, revised and expanded from the 2002 Russian original. EMS Textbooks in Mathematics. European Mathematical Society (EMS), Zürich, 2008. x+177 pp.). </p> <p>More precisely, a finite $p$-group $G$ of order $n$ embeds into $GL=GL_n({\mathbb F}_p)$ (Cayley theorem). The subgroup $UT$ of upper triangular unipotent matrices of $GL$ is a Sylow p-subgroup of $GL$ (proof by computing the order of $GL$), so $G$ is conjugate to a subgroup of $UT$.</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55687#55687 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-17T03:13:33Z 2011-02-17T03:25:32Z <p>I guess you mean finite $p$-groups. Then every finite-dimensional representation of a finite $p$-group $G$ is unipotent in characteristic $p$. (Indeed, all eigenvalues of every element of $G$ are $p$-power roots of unity and therefore must be equal to $1$. This means that every element of $G$ acts as a unipotent operator.) An example of a faithful unipotent representation of $G$ is provided by its regular representation over any field of characteristic $p$. In particular, the regular representation of $G$ over the prime field $F_p$ is faithful and unipotent.</p> <p>(The same construction works for infinite $p$-groups $G$ as well. Of course, in this case the regular" representation space of functions on $G$ with finite support will be infinite-dimensional.)</p> http://mathoverflow.net/questions/55685/faithful-unipotent-representations-of-finite-p-groups/55814#55814 Answer by Yuri Zarhin for faithful unipotent representations of (finite) $p$-groups Yuri Zarhin 2011-02-18T02:44:44Z 2011-02-18T02:44:44Z <p>Here is a slightly different approach. If $G$ is a $p$-group (infinite or finite) and $k$ is a field of characteristic $p$ then in the group algebra $k[G]$ we have $(x-1)^{p^r}=x^{p^r}-1$ for all $x \in k[G]$. (Actually, if elements $a$ and $b$ of any $k$-algebra do commute then $(a-b)^{p^r}=a^{p^r} - b^{p^r}$, thanks to divisibility properties of binomial coefficients.) Applying it to $x=g$ where $g$ is an element of $G$ of order $p^r$, we conclude that $(g-1)^{p^r}=g^{p^r}-1=1-1=0$ in $k[G]$. Since every representation space $V$ of $G$ over $k$ is a module over $k[G]$, we conclude that $g-1$ acts on $V$ as a nilpotent operator.</p> <p>An example of a faithful representation of $G$ is provided by the regular representation where $V$ is the space of all $k$-valued functions on $G$. Another example is provided by its $G$-invariant space $V_0$ of all functions $f: G \to k$ with finite support (i.e., vanishing at all but finitely many points of $G$). Notice that for each $g \in G$ the space $V_0$ splits into a direct sum of $g$-invariant finite-dimensional subspaces (that correspond to finite left cosets of the cyclic group generated by $g$).</p>