What are the normal subgroups of $PSL_2(\mathbb{Z}/p^n \mathbb{Z})$?
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asked Jul 31, 2011 at 23:55
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1$\begingroup$ Pick an $n$ and a $p$ and you can ask GAP to list them for you. Is that all you want -- a list? $\endgroup$– Ryan BudneyCommented Aug 1, 2011 at 0:09
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$\begingroup$ One can think of some obvious ones; presumably you want to know if these are the only ones? $\endgroup$– Yemon ChoiCommented Aug 1, 2011 at 0:39
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1 Answer
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Let's suppose that $p >3$ (otherwise, the groups is solvable in any case). I also work with $G = {\rm SL}(2,\mathbb{Z}/p^n \mathbb{Z})$, but there is an obvious correspondence between what happens for ${\rm PSL}$ and what happens for ${\rm SL}$. The group $G$ has a largest normal $p$-subgroup (denoted, as is customary for group theorists, by $O_p(G)$), and a central subgroup $Z$ of order $2$. $G/Z \times O_{p}(G)$ is isomorphic to the simple group ${\rm PSL}(2,\mathbb{Z}/p\mathbb{Z})$. Furthermore, $H = O_{p}(G)$ is the set of elements of $G$ which are congruent (entrywise) to the identity $(mod p)$. More generally, we have a series of obvious normal subgroups $$ 1 = H_n < H_{n-1} < \ldots < H_{1} = H < G,$$ where $H_{i}$ is the set of elements of $G$ congruent (entrywise) to the identity $(mod p^i)$. For each $j$, $H_{j}/H_{j+1}$ is an elementary Abelian $p$-group (that is, an Abelian $p$-group of exponent $p$). To see this, note that for $a,b \in H_j$, we have $ab -ba = (a-I)(b-I) - (b-I)(a-I) \equiv 0$ (mod $p^{2j})$, so that $I - a^{-1}b^{-1}ab \equiv 0$ (mod $p^{2j}$). Hence $[H_{j},H_{j}] \subseteq H_{2j} \subseteq H_{j+1}$ and $H_{J}/H_{j+1}$ is Abelian (in fact $H_{j}/H_{2j}$ is Abelian). Also , for $a \in H_j$, we have $a^{p} - I = (a-I)(a^{p-1}+ \ldots a^{p-2} + \ldots + I) \equiv 0$ (mod $p^{j+1})$, so that $a^p \in H_{j+1}$. Each of the factor groups $H_{j}/H_{j+1}$ has the structure of a module for the group ${\rm SL}(2,\mathbb{Z}/p\mathbb{Z})$ over the field of $p$-elements.
answered Aug 1, 2011 at 8:17
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$\begingroup$ I guess it should be: "... $H_i$ is the set of elements congruent to the identity $\mod p^i$", and $H= H_1$ in the displayed equation? Alternatively, $H_i$ is the kernel of $ {\rm SL}(2, \mathbb{Z}/p^n \mathbb{Z}) \to {\rm SL} (2,\mathbb{Z}/p^i \mathbb{Z})$. $\endgroup$ Commented Aug 1, 2011 at 10:57
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$\begingroup$ Yes, thanks, I started to write the chain the other way round. Now corrected, I think $\endgroup$ Commented Aug 1, 2011 at 11:08
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$\begingroup$ Thanks Geoff. Is it known whether there are any normal subgroups which aren't contained in H? $\endgroup$ Commented Aug 1, 2011 at 13:01
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1$\begingroup$ @bobgiraffe: For $p>3$, $G={\rm SL}(2, \mathbb{Z}/p^n\mathbb{Z})$ is perfect, if I remember correctly. (I think this can be shown in a similar way as showing that ${\rm SL}(2, \mathbb{F})$ is perfect for fields with more than 3 elements.) Since the other composition factors of $G$ are abelian, it follows that $H$ is the unique maximal normal subgroup of $G$. $\endgroup$ Commented Aug 2, 2011 at 11:20
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1$\begingroup$ @bobgiraffe: for $G= SL_2(\mathbb{Z})$, $G/[G,G]$ is cyclic of order 12. The group $SL_2(\mathbb{Z}/m\mathbb{Z})$ is generated by two elements of order $m$, so commutator factor group of this group is cyclic of order dividing $gcd(m,12)$. If you know, for example, that $SL_2(\mathbb{Z}/4\mathbb{Z})$ has factor group $\mathbb{Z}/4\mathbb{Z}$, then you have an epimorphism $H=SL_2(\mathbb{Z}/2^n \mathbb{Z}) \to SL_2(\mathbb{Z}/4\mathbb{Z}) \to \mathbb{Z}/4\mathbb{Z}$ for $n\geq 2$, which implies $|H/[H,H]|\geq 4$. See math.uconn.edu/~kconrad/blurbs (the note on $SL_2(\mathbb{Z})$). $\endgroup$ Commented Aug 26, 2011 at 12:54