I need to show that $GL(2,\mathbb{Z})=\left\{ \begin{pmatrix} a&b \\ c&d \end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad-bc = \pm 1\right\}$ is not nilpotent. I have seen this question but the answer given there is too advanced for where I am currently in my studies.

In order to show something is not nilpotent, all I have at my disposal is to show that the upper central series does not terminate or that it is not solvable (using a derived series that does not terminate or a subnormal series that either does not terminate or whose factors are not all abelian.)

This seems a little messy, and I've never used a direct approach to show that something was not nilpotent.

What is the best approach to this proof without appealing to $p$-groups, Sylow Groups, Galois Theory, or free abelian groups?

Thank you.

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    $\begingroup$ I mean, what have you tried? The center is pretty easy to find. What does the second center look like? $\endgroup$ – Steve D Jan 9 '17 at 6:28
  • $\begingroup$ @SteveD actually, I lied (okay, I didn't lie - I just hadn't figured it out yet). The center of $GL(2,\mathbb{Z})$ is $\pm$ the identity matrix. Now, I'm guessing that the center of that set is just the identity matrix. $\endgroup$ – ALannister Jan 9 '17 at 22:02
  • $\begingroup$ @SteveD again, lies. I meant that the center of the center of $GL(2,\mathbb{Z})$ is all of $GL(2,\mathbb{Z})$. Is this what you meant by the "second center"? Oh, I hope you respond, because I've found this hint more helpful than either of the two answers below! $\endgroup$ – ALannister Jan 9 '17 at 22:39

Let's find the upper central series. For ease of notation, let $$ T_k = \begin{pmatrix} 1 & k\\0 & 1\end{pmatrix}$$ and $$ R = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$$ be translation and reflection matrices, with inverses $T_k^{-1} = T_{-k}$ and $R^{-1}=R$. The center of $GL(2,\mathbb{Z}$) must surely commute with all $T_k$ (that is, for all $k$), and also with $R$. If we let $$ X = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$$ then $X$ being in the center implies \begin{align} T_kXT_{-k} &= X\\ RXR &= X \end{align} This is enough to imply $b=c=0$ and $a=d$ [to see this, just do the multiplcations and equate matrix coefficients]; that is, the center is $Z_1=\{I, -I\}$, with $I$ the $2\times2$ identity matrix.

What about the second center, $Z_2$? These are elements that are central "mod $Z_1$", or elements $X$ such that for all $Y\in GL(2,\mathbb{Z})$, we have $$ YXY^{-1} = \pm X$$ [Note: the choice of $+$ or $-$ on the right hand side is dependent on $Y$.]

Again, using $T_k$ and $R$, show that $X$ would still have to be $\pm I$. This means $Z_2=Z_1$, and thus our upper central series has stalled (and hasn't become the whole group); the group $GL(2,\mathbb{Z})$ thus is not nilpotent.

EDIT: Adding in more details.

First center

If $X$ is in the center of $GL(2,\mathbb{Z})$, then it commutes with all matrices. In particular, we have $$ T_kXT_{-k}=X$$ for all $T_k$. Expanding that out, we get four equations: \begin{align} a+ck &= a\\ b+k(d-a-ck) &= b\\ c &= c\\ d-ck &= d \end{align} Since this holds for any $k$, we get from the first equation $a+ck_1=a+ck_2$ for $k_1\neq k_2$, which immediately implies $c=0$. Knowing $c=0$, the second equation similarly implies $a=d$. To get $b=c$, we expand $RXR=X$: \begin{equation} \begin{pmatrix} d & c\\ b & a \end{pmatrix} = \begin{pmatrix} a & b\\c & d\end{pmatrix} \end{equation}

Thus, our center is $\{I, -I\}$ (we actually need to check these commute with all matrices, but that's easy).

Second center

The second center is defined as $Z_2(G)/Z_1(G)=Z(G/Z_1(G))$; that is, it consists of the matrices that commute with all others, "mod $Z_1$". This means that if $X\in Z_2$, and $Y\in GL(2,\mathbb{Z})$, then there exists $Z_Y\in Z_1$ such that $$ YXY^{-1} = Z_YX$$ The element from $Z_1$ will probably depend on $Y$; the point is that it always exists. (If you want to think about this more abstractly, the above is just saying that $[Z_2, G]\subset Z_1$).

Let's find the second center. If $X\in Z_2$, then for every $k$, there exists $Z_k\in Z_1=\{I, -I\}$ such that $$ T_kXT_{-k} = Z_kX$$

Since there are only two choices for $Z_k$ (namely, $I$ or $-I$), and since there are infinitely many $k$s, we can choose $k_1, k_2$ such that $Z_{k_1}=Z_{k_2}$, and thus just as above, we get (after possibly cancelling a negative sign if $Z_{k_1}=Z_{k_2}=-I$) $$ a + ck_1 = a + ck_2$$ which again implies $c=0$. Using $R$ we have \begin{equation} \begin{pmatrix} d & c\\ b & a \end{pmatrix} = Z_R\begin{pmatrix} a & b\\c & d\end{pmatrix} \end{equation} where $Z_R$ is $I$ or $-I$. Thus we get $b=\pm c=0$, and $a=\pm d$.

That means the only thing left to show that $Z_2=Z_1=\{I, -I\}$ is to show that $a=-d$ cannot happen. That is, we need to show the matrix $$ \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix} $$ is not in $Z_2$. I'll leave this as an exercise (hint: try checking what happens with $T_1$).

  • $\begingroup$ because in the proof for $RXR=\pm X$, we have that $\begin{pmatrix} d&c \\ b & a \end{pmatrix} = \pm \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, which implies, in the case that $\begin{pmatrix} d&c \\ b & a \end{pmatrix} = -\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, that $d=-a$, $c=-b$, which doesn't make sense, since $X$ is supposed to be one of $I$ or $-I$. Even if you don't reply to any of my other stupid comments, reply to this one please. I'd like to accept your answer, but I can't do it if I'm not 100% convinced it's correct. $\endgroup$ – ALannister Jan 10 '17 at 2:54
  • $\begingroup$ please. I don't understand this $\pm$ part I alluded to in the last comment. Please either reply to my comment or edit your answer to make it clearer. $\endgroup$ – ALannister Jan 10 '17 at 3:16
  • $\begingroup$ Seems like it should be $\pm X$ on BOTH sides of the equals sign @SteveD $\endgroup$ – ALannister Jan 10 '17 at 3:30
  • $\begingroup$ @JessyCat: saying that $X$ is in the second center, is just saying that $X$ commutes with every element, up to multiplication by an element in the first center. That is, if $X\in Z_2$ and $Y$ is any other element, then there exists $Z_Y \in Z_1$ [the regular center] such that $YXY^{-1}=Z_YX$. This $Z_Y$ might depend on $Y$, but it's always in the center. Since $Z_1=\{I, -I\}$, that's where the plus/minus comes from. $\endgroup$ – Steve D Jan 10 '17 at 3:34
  • $\begingroup$ @JessyCat: $X$ is not necessarily $I$ or $-I$. It's some arbitrary element; we're trying to show that if we assume $X\in Z_2$, then it is $I$ or $-I$. Note that in your first comment, it Does make sense, since we are only working up to $\pm$. $\endgroup$ – Steve D Jan 10 '17 at 3:35

One possibility is to show that there is a proper, self-normalizing subgroup, $$ \left\{ \begin{pmatrix} a&b \\ 0&d \end{pmatrix} : a, d = \pm 1, b \in \mathbb{Z}\right\} $$ being perhaps the easiest candidate.

  • $\begingroup$ I've never come across the idea of a self-normalizer before, so probably not basic enough for me. Thanks anyway, though. $\endgroup$ – ALannister Jan 9 '17 at 14:33
  • $\begingroup$ I just mean that the normalizer of the subgroup in the whole group is the subgroup itself. $\endgroup$ – Andreas Caranti Jan 9 '17 at 15:33
  • $\begingroup$ okay, but why does that imply that the whole group is not nilpotent? I would need to prove such a result before I could use it. $\endgroup$ – ALannister Jan 9 '17 at 15:48
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    $\begingroup$ Right. If $G$ is nilpotent, and $H < G$ is a proper subgroup, then $H < N_{G}(H)$. This is proved by taking the upper central series $1 = Z_{0} < Z_{1} = Z(G) < Z_{2} < \dots < Z_{n} = G$. If $i$ is such that $Z_{i-1} \le H$ and $Z_{i} \not\le H$, then $H < Z_{i} H \le N_{G}(H)$, as $[Z_{i} H, H] \le [Z_{i}, H] H \le Z_{i-1} H = H$. $\endgroup$ – Andreas Caranti Jan 9 '17 at 16:11

Another possibility is to show that the group has a non-nilpotent homomorphic image. Consider the reduction modulo two mapping, and identify $GL_2(\Bbb{F}_2)$ as the smallest non-nilpotent group.

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    $\begingroup$ +1, good idea. Possibly a simple proof that the group is not even soluble would be best in some sense. (Of course anything can be conceived as best in some [perverted] sense.) However, this is probably not so short. $\endgroup$ – Andreas Caranti Jan 9 '17 at 10:20
  • $\begingroup$ @JyrkiLahtonen would you mind explaining what $GL_{2}(\mathbb{F}_{2})$ is? $\endgroup$ – ALannister Jan 9 '17 at 14:35
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    $\begingroup$ @JessyCat The set of invertible 2x2-matrices with entries in the field $\Bbb{F}_2\simeq\Bbb{Z}/2\Bbb{Z}$. Counting will reveal that there are six such matrices. The group they form is not abelian. $\endgroup$ – Jyrki Lahtonen Jan 9 '17 at 14:37
  • $\begingroup$ @JyrkiLahtonen oh, ok. Thank you. I know what $\mathbb{Z}/2\mathbb{Z}$ is, but I've never heard of it being called "$\mathbb{F}_{2}$" before. $\endgroup$ – ALannister Jan 9 '17 at 15:17
  • $\begingroup$ @AndreasCaranti: As you probably know, if you look at the image in $GL_2(F_p)$ you get something which is almost non-abelian simple (i.e. simple up to obvious abelian subs. and quotients) when $p \geq 5$, since it has the simple group $PSL_2(F_p)$ as a subquotient. Probably the easiest prime to deal with is $p = 5$, when this is $A_5$; of course you have to prove it, but I guess this is doable, e.g. by finding a concrete subgroup of index 5. (Not that I'm proposing this as the best way to answer the OP's question, but it might be one of the quicker routes to non-solvability.) $\endgroup$ – tracing Jan 9 '17 at 23:56

Because I'm lazy, I'll write $G = \operatorname{GL}(2,\mathbb{Z})$.

From your comments/messages in the chat, I gather that you found the center $Z(G) = \{I,-I\}$ where $I$ is the identity matrix.

Now, consider the upper central series: $$1 = Z_0 \trianglelefteq Z_1 \trianglelefteq Z_2 \trianglelefteq \cdots \trianglelefteq Z_i \trianglelefteq \cdots$$ where $$Z_{i+1} = \{ x \in G \mid \forall y \in G: [x,y] \in Z_i\}= \{ x \in G \mid \forall y \in G: x^{-1}y^{-1}xy \in Z_i\}.$$ In particular, observe that $$Z_1 = \{ x \in G \mid \forall y \in G: [x,y] \in Z_0\} = \{ x \in G \mid \forall y \in G: xy = yx \} = Z(G) = \{I,-I\}.$$ Now, onwards to $Z_2$. $$Z_2 = \{ x \in G \mid \forall y \in G: [x,y] \in Z_1\} = \{ x \in G \mid \forall y \in G: (xy = yx \text{ or } xy = -yx) \}.$$ Let us examine the second condition: $xy = -yx$. This means that $y = -x^{-1}yx$, which gives $$\operatorname{trace}(y) = \operatorname{trace}(-x^{-1}yx) = -\operatorname{trace}(x^{-1}yx) = -\operatorname{trace}(xx^{-1}y) = -\operatorname{trace}(y)$$ hence $\operatorname{trace}(y) = 0$. So $x$ must commute with all matrices $y$ having $\operatorname{trace}(y) \neq 0$. We claim that the only matrices $x$ for which this holds, are $I$ and $-I$ (see below for a proof of this claim).

Thus we end up with $Z_2(G) = Z(G)$ again! So if we repeat these steps again, we'll find that $Z_3 = Z(G)$, and again that $Z_4 = Z(G)$, etc: $Z_i = Z(G) = \{I,-I\}$ for all $i \geq 1$.

But then the upper central series never terminates; and hence $G$ is not nilpotent!

Proof of claim: Clearly it holds for $x = I$ or $x = -I$, since these groups are in the center of $G$.

Now, let $x = \begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix}$. Then we must have that $$\begin{pmatrix}x_1 & x_1+x_2\\x_3 & x_3 + x_4\end{pmatrix} = \begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix}\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix}x_1 & x_2\\x_3 & x_4\end{pmatrix} = \begin{pmatrix}x_1 +x_3& x_2+x_4\\x_3 & x_4\end{pmatrix},$$ hence $x_3 = 0$ and $x_1 = x_4$. Similarly, commuting $x$ with $\begin{pmatrix}1 & 0\\ 1 & 1\end{pmatrix}$ you find that $x_2 = 0$ and $x_1 = x_4$. But since $x \in G$, then the only matrices satisfying this are $I$ and $-I$.

  • $\begingroup$ The condition $xy=yx$ or $xy=-yx$ is dependent on $y$; it is not the case that it must "always be positive" or "always be negative". $\endgroup$ – Steve D Jan 10 '17 at 0:49

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