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(Def): Let $G$ be a group and $X \subset G$. Let $\{ H_{\alpha} \}_{\alpha \in \Gamma} $ be a collection of all subgroups of $G$ which contain $X$. Then $\bigcap_{\alpha \in \Gamma} H_{\alpha} $ is the subgroup of $G$ generated by the set $X$ and is denoted by $\langle X \rangle $.

(Problem):

If $G$ is a group and $X$ is a nonempty subset of $G$, then $\langle X \rangle $ consists of all finite products

$$ a_1^{n_1} a_2^{n_2} ... a_t^{n_t} $$

for all $a_i \in X$ and $n_i \in \mathbb{Z} $. In particular, for all $a \in G $ we have $\langle a \rangle = \{a^n : n \in \mathbb{Z} \} $

Attempt

Let $H = \{ a_1^{n_1} a_2^{n_2} ... a_t^{n_t} \} $. Notice that $H$ is a subgroup of $G$ since if we take $h,g \in H$, then $hg^{-1}$ will lie in $H$. Since any element in $X$ can be expressed in the form of the elements of the set $H$, then we have $X \subset H$. In particular, $H$ is part of the collection of subgroups of $G$ that contain $X$. Hence,

$$ \langle X \rangle = \bigcap H_{\alpha} \subset H$$

Next, if we take $h \in H$, then $h \in X \subset H_{\alpha} $ for all $\alpha$. Hence, $H$ is contained in all the $H_{\alpha} 's$. In particular,

$$ H \subset \bigcap H_{\alpha} = \langle X \rangle $$

Therefore $H = \langle X \rangle $, as desired.

Question: Is this a correct approach? A feedback would be extremely thankful.

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  • $\begingroup$ $a_i^{n_i} \notin X$, in general. your idea is fine, but you made some mistakes by assuming that $a_i^{n_i} \in X.$ try it again, you can prove it. $\endgroup$ – Krish Nov 23 '14 at 6:00
  • $\begingroup$ $H = \{a_1^{n_1}\dots a_t^{n_t}\}$, so $H$ has one element? $a_i^{n_i}$ is in $X$? $hg^{-1}$ will lie in $H$? None of these statements seem accurate, relevant, or justified. Maybe you can start by being more explicit. Be clear about what you're trying to prove, why you're trying to prove it, what deductions you're making along the way, and what justifies those deductions. $\endgroup$ – Amit Kumar Gupta Nov 23 '14 at 6:03
  • $\begingroup$ I edited my answer. $hg^{-1}$ will also be a product of elements of $X$, so it will live in $H$. $\endgroup$ – ILoveMath Nov 23 '14 at 6:11
  • $\begingroup$ @FromCuba: you should write more clearly as Amit Kumar Gupta pointed out. for example, $H := \{a_1^{n_1} a_2^{n_2} ... a_t^{n_t} | a_i \in X, n_i \in \mathbb{Z} \}.$ $\endgroup$ – Krish Nov 23 '14 at 6:31
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$h \in H$ does not imply that $h \in X$. On the other hand, $h \in H \Rightarrow h = a_1^{n_1} a_2^{n_2} ... a_t^{n_t},$ for some $a_1, a_2, \cdots, a_t \in X$ and $n_1, n_2, \cdots , n_t \in \mathbb{Z}$. If $H_{\alpha}$ is any subgroup of $G$ containing $X$, then $a_i^{n_i} \in H_{\alpha}$ for each $i$ and so $h \in H_{\alpha}.$ Thus $H \subseteq \bigcap_{\alpha \in \Gamma}H_{\alpha}.$

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  • $\begingroup$ I understand now. The other direction (the other inclusion) of my solution is correct? $\endgroup$ – ILoveMath Nov 23 '14 at 6:34
  • $\begingroup$ yes. it's fine. $\endgroup$ – Krish Nov 23 '14 at 6:37
  • $\begingroup$ Thank you very much for your time my friend. I really appreciate it =) . $\endgroup$ – ILoveMath Nov 23 '14 at 6:38
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Start by defining $H$ to be exactly the set given: $$H = \left\{a_1^{n_1} a_2^{n_2} \ldots a_k^{n_t} \mid t \in \mathbb{N}, a_1,\ldots,a_t \in X, n_1,\ldots,n_t \in \mathbb{Z}\right\}$$

Then show that there indeed $a,b \in H$ then $ab^{-1} \in H$, writing down the explicit formula. Then $\left\langle X \right\rangle \subseteq H$ is true, as you proved. (Also note that $x \in X$ is just written as $x^1$ so in $H$ by definition.)

On the other hand, if $H_i$ is any subgroup of $G$ that contains $X$, it must contains all members of $H$ by virtue of containing $X$ and being closed under powers. So $H \subseteq H_i$, and so the other inclusion follows too.

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