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- #1

- Mar 10, 2012

- 835

I am stuck at this problem for quite a long time now.

**Let $F_p$ denote the field of $p$ elements, where $p$ is prime. Let $n$ be a positive integer. Let $V$ be the vector space $(F_p)^n$ over the field $F_p$. Let $GL_n(F_p)$ denote the set of all the invertible linear transformations on $V$. Then note that each element $T\in GL_n(F_p)$ is a permutation of the elements in $V$. Show that there is an element $T\in GL_n(F_p)$ such that the cycle decomposition of $T$ has a cycle of length $p^n-1$.**

Problem.

Problem.

**Attempt.**

Write $s=p^n=|V|$. Let $T\in\mathcal L(V)$ and $v\in V$. We say that $T$ is good with $v$ if $\{Tv,\ldots,T^{s-1}(v)\}=V\setminus\{0\}$.

*Claim 1:*If $T\in \mathcal L(V)$ is good with a vector $v\in V$ then $T\in GL_n(F_p)$.

*Proof:*Let $K=\{v,Tv,\ldots,T^{s-2}(v)\}$. Then $T(K)=V\setminus\{0\}$ since $T$ is good with $v$. Thus $T$ is surjective and hence invertible.

*Claim 2:*If $T$ is good with a vector $v\in V$ then $T^kv=v\Rightarrow k\geq s$.

*Proof:*Let $k$ be the smallest positive integer with the property that $T^kv=v$. Now $\{Tv,\ldots,T^{s-1}v\}=\{Tv,\ldots,T^{k-1}v\}$. The LHS has cardinality $s-1$ while the RHS has the cardinality $k-1$. Hence $s=k$ and we are done.

*Claim 3:*If $T$ is good with a vector $v$ then $T^sv=v$.

*Proof:*Clearly $T^sv\in V\setminus\{0\}$. So let $T^sv=T^iv$ for some $i$ in $\{1,\ldots,s-1\}$. This gives $T^{s-i}v=v$ and form the above claim $s-i\geq s$ and hence $i\leq 0$, which is a contradiction.

*Claim 4:*If $T$ is good with $v\in V$ then $T$ is good with each $u\in V\setminus\{0\}$.

*Proof:*Let $u\in V\setminus\{0\}$. Since $T$ is good with $v$, we have $u=T^iv$ for some $i \in \{1,\ldots,s-1\}$. Let $T^ku=u$ and $k$ be minimum, positive integer with this property. Then $T^{i+k}v=T^{i}v$. This leads to $T^k v=v$ and hence $k\geq s$. This settles that $T$ is good with $u$.

*Claim 5:*Let $T\in \mathcal L(V)$ is good with a vector $v\in V$ if and only if $T^k-I$ is invertible for each $k\in\{1,\ldots,s-1\}$.

*Proof:*Let $T$ be good with a vector $v\in V$ but $T^k-I$ is not invertible for a $k\in\{1,\ldots,s-1\}$. Then there is a vector $u\in V\setminus\{0\}$ such that $(T^k-I)u=0$. This means $T^ku=u$ and hence $T$ is not good with $u$. But this implies that $T$ isn't good with $v$ either and hence this direction of the claim in settled. The other direction is trivial.

Now coming back to the problem.

The question asks us to establish that there is some $T\in GL_n(F_p)$ such that $T$ is good with some vector $v\in V$. By Claim 5 this is equivalent to showing that there is a $T$ in $\mathcal L(V)$ such that $T^k-I$ is invertible for each $k\in\{1,\ldots,s-1\}$. To do this may be it will help to use the fact that an operator is invertible if and only if its matrix with respect to any basis has determinant non zero. I am unable to make progress here. Can anybody help?