Roots of unity in GL(2) modulo center

Question

I am aware of that fact that if $p>n+1$, then there are no elements of order $p$ in $GL_n(\mathbb{Z}_p)$, where $\mathbb{Z}_p$ is the ring of $p$-adic integers. I want to show that the same holds for $PGL_n(\mathbb{Z}_p)$ which is $GL_n(\mathbb{Z}_p)$ modulo its center. Any help will be appreciated.

Proof in the $GL_n$ case is as follows:

For $ X \in GL_n(\mathbb{Z}_p)$, suppose $X^p=1$. Then, $X^p-1=(X-1)(X^{p-1}+\cdots+1)$ and $(X^{p-1}+\cdots+1)$ is irreducible over $\mathbb{Z}_p$. So the minimal polynomial of $X$ is of degree $p-1$. But the characteristic polynomial of $X$ is of degree $n$ and hence $n \geq p-1$.

But things are getting complicated because of the center. Any help or reference?

score 4 · Accepted Answer · answered Jun 25 '21 at 14:39

4

The argument essential works the same way. Let's use the fact that $\mathrm{GL}_n(\mathbb{Z}_p)\to \mathrm{PGL}_n(\mathbb{Z}_p)$ is surjective.

Suppose $X\in \mathrm{PGL}_n(\mathbb{Z}_p)$ satisfies $X^p=I_n$, and let $Y\in \mathrm{GL}_n(\mathbb{Z}_p)$ be a lift of $X$, which always exists by surjectivity. Then there exists $a\in \mathbb{Z}_p$ such that $Y^p=aI_n$ lies in the center. This implies that the minimal polynomial of $Y$ divides $t^p-a$. If $a$ is not a $p^{th}$ power, this polynomial is irreducible, and as in your case we obtain $n\geq p$.

If $a=b^p$ for some $b\in \mathbb{Z}^\times_p$, then $$t^p-b^p=(t-b)(t^{p-1}+bt^{p-2}+\cdots +b^{p-2}t+b^{p-1}).$$ If the minimal polynomial divides $t-b$, then $Y=bI_n$ so that $X=I$ in the quotient. Otherwise the minimal poly divides the second factor, which we claim is also irreducible. In particular, we get $n\geq p-1$ in this case. The upshot is that as soon as $p>n+1$, you get the result.

To see that the polynomial $g(t)=t^{p-1}+bt^{p-2}+\cdots +b^{p-2}t+b^{p-1}$ is irreducible over $\mathbb{Z}_p$, suppose it factors as $g(t) = p(t)q(t)$ with both factors having positive degree. Then $$ b^{p-1}(t^{p-1}+t^{p-2}+\cdots +t+ 1) = g(bt) = p(bt)q(bt)$$ gives a non-trivial factorization of $t^{p-1}+t^{p-2}+\cdots +t+ 1$ in $\mathbb{Z}_p[t]$, a contradiction.

answered Jun 25 '21 at 14:39

Spencer Leslie

2,662

Thanks Spencer. If a is a $p^$th power. Let $a=b^p$ and $b$ is invertible and lies in the center. Then $Y^p-a=0$ is the same as $(b^{-1}Y)^p-1=0$. So $b^{-1}Y$ is an element of order $p$ in $GL_n$. So $b^{-1}Y=1$ applying the result for GL(n). But then $Y=b$ and so $Y$ is trivial. Is this also correct? – Questions-Math Jun 25 '21 at 15:05
If $a$ is not a power of $p$, why is $t^p-a$ irreducible? – Questions-Math Jun 25 '21 at 15:06
This is a more general algebra fact; see the discussion at https://math.stackexchange.com/questions/403924/xp-c-has-no-root-in-a-field-f-if-and-only-if-xp-c-is-irreducible?noredirect=1&lq=1 – Spencer Leslie Jun 25 '21 at 15:11
For the first comment, yes that works too. I opted to just play with the polynomials :) – Spencer Leslie Jun 25 '21 at 15:13
Thanks so much Spencer – Questions-Math Jun 25 '21 at 15:22
Happy to help!! – Spencer Leslie Jun 25 '21 at 15:35

Roots of unity in GL(2) modulo center

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