Let $G$ be a finite group and $H$ a subgroup of $G$. There exists a sequence of cyclic subgroups $H_1, ..., H_n$, not necessarily distinct, such that $H$ is the product $H = H_1H_2\cdots H_n = \{ h_1h_2\cdots h_n : h_i\in H_i,\forall i\}$.

  • 1
    $\begingroup$ I don't see why arbitrary finite groups could be written as products of cyclic groups (and obviously internal direct products would be false), nor how this could be used to find subgroups of small groups. Every group is generated by cyclic groups, though. $\endgroup$ – anon Aug 6 '13 at 3:06
  • $\begingroup$ @anon I don't think OP means direct product, but rather "subgroup products." $\endgroup$ – Thomas Andrews Aug 6 '13 at 3:08
  • $\begingroup$ How are you defining $H_1H_2$? Is it the smallest subgroup contain $H_1$ and $H_2$, or is it the set $\{h_1h_2:h_1\in H_1, h_2\in H_2\}$? The latter is not a subgroup. $\endgroup$ – Thomas Andrews Aug 6 '13 at 3:11
  • $\begingroup$ @ThomasAndrews That's why I distinguished the two. $\endgroup$ – anon Aug 6 '13 at 3:12
  • 3
    $\begingroup$ Just take all cyclic subgroups of $H$... $\endgroup$ – Mariano Suárez-Álvarez Aug 6 '13 at 3:22

Let $H_1$, $\dots$, $H_n$ be the list of all cyclic subgroups of $H$. Then $H=H_1\dots H_n$.

Indeed, every element $x$ of $H$ belongs to one of the $H_i$ (for example, to the cyclic group it generates), and if $H_k$ is one which contains it then $$x=\underbrace{e\cdot e\cdot\cdots\cdot e}_{\text{$k-1$ factors}}\cdot x\cdot \underbrace{e\cdot\cdots\cdot e}_{\text{$n-k$ factors}}\in H_1\cdots H_n.$$

  • $\begingroup$ Cool, i see how that works. $\endgroup$ – BananaCats Author Aug 6 '13 at 3:24
  • $\begingroup$ Can you drop to the chat, Mariano? $\endgroup$ – Pedro Tamaroff Aug 6 '13 at 3:36

This is true, but I don't know how useful it's going to be, because the cyclic subgroups of $G$ aren't enough to characterize the structure of $G$. For example, the modular group of order 16, $$M_{16}=\langle a,b~|~a^2=b^8=1,ba=ab^5\rangle,$$ and the product of cyclic groups of orders 2 and 8, $$C_2\times C_8 = \langle a,b~|~a^2=b^8=1, ba=ab\rangle,$$ have the same cycle graph.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.