On abelian subgroup of $GL_{n}(\mathbb{F}_{p})$. Let $p$ be a prime number, and let $\mathbb{F}_{p}$ be a finite
field of order $p$. Let $G=GL_{n}(\mathbb{F}_{p})$ denote the general linear
group and $U_{n}$ denote the unitriangular group of $n\times n$ upper
triangular matrices with ones on the diagonal, over the finite field $%
\mathbb{F}_{p}$. Let $H$ be an abelian subgroup of order $p^{m}$ in $U_{n}$. Does the subgroup $H$ must be elementary abelian of rank $m$ ( $H\simeq (\mathbb{F}_{p})^{m})$. 
Any help would be appreciated so much. Thank you all.  
 A: The answer is no for $n=4$, $p=3$ and $m=2$. The group $UT(4,3)$ has a cyclic subgroup of order $9$.
A: Assume $\mathcal H$ is an abelian $p$ subgroup of upper unitriangular matrices in $\text{GL}_n(\Bbb F_p)$ and choose $\mathbf T:=(t_{ab})_{1\leq a,b\leq n}\in\mathcal H$. Choose any $k,1< k\leq n,$ satisfying $\mathbf T^p\cdot\vec{\mathbf v}_j=\vec{\mathbf v}_j$ for all $j<k$ where $$\vec{\mathbf v}_j:=(t_{1j},\dots,t_{jj},0,\dots,0)=t_{1j}\mathbf e_1+\cdots +t_{jj}\mathbf e_j$$ is column $j$ of $\mathbf T$ and $\mathbf e_j$ denotes column $j$ of the identity matrix. Consequently, we should also have $\mathbf T^p\cdot\mathbf e_j=\mathbf e_j$ for all $j<k$ . 
Therefore, $$\mathbf T^p\cdot\vec{\mathbf v}_k=\mathbf T^p\cdot (t_{1k}\mathbf e_1+\cdots +t_{kk}\mathbf e_k)=\mathbf T^p\cdot\mathbf e_k+\sum_{r=0}^{p-1}\mathbf T^r\cdot(\vec{\mathbf v}_k-\vec{\mathbf e}_k)=\mathbf T^p\cdot\mathbf e_k+(\sum_{r=0}^{p-1}\mathbf T^r)\cdot(\vec{\mathbf v}_k-\vec{\mathbf e}_k)=\vec{\mathbf v}_k$$ 
By induction we must have $\mathbf T^p\cdot\vec{\mathbf v}_j=\vec{\mathbf v}_j$ for all $j$ and therefore $\mathbf T^p=\mathbf I$ is the identity matrix. Consequently, $\mathbf X^p=\mathbf I$ for all $\mathbf X\in\mathcal H$. Therefore, from the fundamental theorem of finitely generated abelian groups we must have $\mathcal H\approx\Bbb F_p^{+m}=\Bbb F_p^+\times\cdots\times\Bbb F_p^+$. 
