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I'm solving this question in Hungerford's book:

8​. Let $G$ be the multiplicative group generated by the real matrices $a=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ and $b=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. If $H$ is the set of all matrices in $G$ whose (main) diagonal entries are $1$, then $H$ is a subgroup that is not finitely generated.

I've already proved that $H$ is a subgroup of $G$ and every element of $H$ is of the form:

$$ \begin{pmatrix} 1 & x \\ 0 & 1 \\ \end{pmatrix} $$

I thought this question really weird, $H$ is a subgroup of a group generated by $a$ and $b$ but $H$ itself can't be generated by $a$ or $b$, how can be possible?

My question is how to show $H$ is infinitely generated? I didn't find any good candidates to generate $H$.

Thanks in advance

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  • $\begingroup$ Re: "really weird", perhaps your idea of generators is not quite accurate. Note that to say $H$ is generated by $a$ and $b$ means, roughly, that calculating all possible combinations of $a$ and $b$ will give you $H$ exactly, that is, $H$ and nothing more. But in your question, $a$ and $b$ generate $G$ which is more than $H$, not $H$ exactly. Not sure whether or not this is the issue which is confusing you but hope it helps. $\endgroup$ Commented Jun 4, 2014 at 7:09
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    $\begingroup$ More importantly, $a$ and $b$ don't lie in $H$, so that they don't generate $H$. By the way, the description of $H$ is incomplete: What is $x$? Not every real number ... $\endgroup$ Commented Jun 4, 2014 at 7:12
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    $\begingroup$ @ivanpenev I doubt $H$ is generated by $b$. For $a^{-1}ba = \left(\begin{matrix}1 & 1/2\\0 & 1\end{matrix}\right) \in H$ (as its diagonal entries are $1$), but it is not generated by $b$ ($b^{-1} = \left(\begin{matrix}1 & -1\\0 & 1\end{matrix}\right)$). $\endgroup$ Commented Jun 4, 2014 at 7:39
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    $\begingroup$ One way to show that $H$ is not finitely generated is to show that it is isomorphic to a group well known not to be finitely generated. Are you familiar with the quasicyclic groups? (Alternatively, you can show directly that any finite set in $H$ must be contained in a proper (in this case, cyclic) subgroup.) $\endgroup$ Commented Jun 4, 2014 at 7:52
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    $\begingroup$ @user42912 Have you considered Martin Brandenburg's hint? Do you know what the values of $x$ are? If not, that should be the next item on your agenda. You need to find out what $H$ is before you try to prove anything more complicated about $H$. $\endgroup$ Commented Jun 4, 2014 at 8:20

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Notice that $H$ is an Abelian group, and all its non-identity elements have infinite order. If it were finitely generated, it would be a direct sum of a finite number of copies of $\mathbb{Z},$ say generated by $\{ \left(\begin{array}{clcr} 1 & x_{i}\\0 & 1 \end{array}\right) : 1 \leq i \leq n \}.$ then every element of $H$ could be expressed (uniquely, though that is not essential here) in the form $\left(\begin{array}{clcr} 1 & u \\0 & 1 \end{array}\right)$, where $u = \sum_{i=1}^{n} z_{i}x_{i},$ with each $z_{i} \in \mathbb{Z}.$ Now the $(1,2)$-entry of any $h \in H$ is certainly a rational number ( in fact with denominator a power of $2$). But the denominator of every $\mathbb{Z}$-combination of the $x_{i}$ is at worst the lcm of the denominators of the $x_{i},$ so is bounded. However, $\left(\begin{array}{clcr} 1 & 2^{-i}\\0 & 1 \end{array}\right)$ lies in $H$ for any $i \in \mathbb{N},$ a contradiction.

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  • $\begingroup$ You said: "If it were finitely generated, it would be a direct sum of a finite number of copies of $\mathbb{Z},$ ", why? $\endgroup$ Commented Jun 5, 2014 at 5:19
  • $\begingroup$ This is the fundamental theorem of Abelian groups, which you may not have seen yet. That theorem says that a finitely generated Abelian group is isomorphic to a direct sum of cyclic groups. It's a special case of the theory of finitely generated modules over a Principal Ideal domain, which you may not have seen. But for the argument above, I don't really need to use that fact that the $x_{i}$ form a basis- it's enough that they generate, the denominator argument still works- and we were assuming $H$$ finitely generated in the first place. $\endgroup$ Commented Jun 5, 2014 at 10:17
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Given $s, t \in \mathbb{Z}$, define $$ h(s,t) := a^{s} b^ta^{-s} = \begin{pmatrix} 1 & 2^s t \\ 0 & 1\end{pmatrix} \in H.$$ Observe that $h(0,0) = 1$, and that for all $s,t,u,v \in \mathbb{Z}$, we have $$ (a^sb^t)(a^ub^v) = h(s',t')a^{s+u}, \quad \text{where} \quad s' = \min\{s,s+u\}, \ t ' = 2^{s-s'}t + 2^{s+u-s'}v.$$ Since each element of $G$ is a product of matrices of the form $a^mb^n$ with $m,n \in \mathbb{Z}$, the above relations imply that each matrix $g \in G$ can be represented in the form $$g = h(s,t)a^r \quad \text{with} \quad s,t,r \in \mathbb{Z}.$$ Note that for $g \notin \langle a\rangle$, this factorisation can be made unique if we require that $t$ be an odd integer. A matrix $g \in G$ belongs to $H$ if and only if $\det(g) = 2^r = 1$, or in other words, if $g = h(s,t)$ for some $s, t\in\mathbb{Z}$.

For $g = h(s,t) \in H$ such that $g \neq 1$, define $\operatorname{ord}(g)$ to be the unique $s' \in \mathbb{Z}$ such that $g = h(s',t')$ with an odd integer $t'$; for $g = 1$ put $\operatorname{ord}(g) = \infty$. Thus, we have $\operatorname{ord} h(s,t) \geq s$ for all $s,t\in\mathbb{Z}$. Now, suppose that we are given $k \geq 1$ pairs of integers $(s_1,t_1),\ldots,(s_k,t_k)$ with $t_j \equiv 1 \!\pmod 2$ for $1 \leq j \leq k$, and let $s = \min\{s_1,\ldots,s_k\}$. For any $k$-tuple $(m_1,\ldots,m_k) \in \mathbb{Z}^k$, we have $$ h(s_1,t_1)^{m_1}\cdots h(s_k,t_k)^{m_k} = h(s, \,2^{s_1-s}m_1t_1 + \cdots + 2^{s_k-s}m_kt_k).$$ Therefore, $\operatorname{ord}\left(h(s_1,t_1)^{m_1}\cdots h(s_k,t_k)^{m_k}\right) \geq s$ for all $(m_1,\ldots,m_k) \in \mathbb{Z}^k$. On the other hand, $\operatorname{ord} h(s-1,1) = s-1$, and therefore, $h(s-1,1) \in H$ does not belong to the group generated by $h(s_1,t_1), \ldots, h(s_k,t_k)$. Hence, $H$ is not finitely generated.

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