0
$\begingroup$

By adding an element $g_j$ to a subgroup $H_{j-1}=\{g_1,g_2,\ldots g_{j-1}\}$, do we get the subgroup $H_j=\langle g \rangle H_{j-1}$? I dont understand how the new group is simply a cyclic extension of the original subgroup, could someone explain? or if answer is a no, what is the basic meaning of $\langle g \rangle H$? Note that $H_{j+1}/H_{j}$ is a cyclic factor group. Also could we say anything about the order of $g_j$?

$\endgroup$
2
  • 1
    $\begingroup$ You can't just add an element to a subgroup and expect it to still be a subgroup. I've been attempting an answer to this question but I think you've been too vague here. Is there any more information you can provide? $\endgroup$ – Ian Coley Mar 31 '13 at 19:13
  • $\begingroup$ We know that $H_j/H_{j-1}$ is an cyclic abelian factor group. Could anything be inferred? $\endgroup$ – Annonymous Mar 31 '13 at 19:17
6
$\begingroup$

If you know that $H_{j+1}/H_j$ is a cyclic factor group, then this is pretty much the definition of $H_{j+1}$ being a cyclic extension of $H_j$.

In general, $G$ is an extension of $K$ by $H$ if $K$ is a normal subgroup of $G$ and $G/K\cong H$. 'Cyclic extension' just means that $K$ is cyclic.

As for the meaning of $\langle g\rangle H$, it just means all elements of the form $g^n h$, where $n\in \mathbb{Z}$ and $h\in H$. In general this might not even be a group, but since you know $H_j$ is normal in $H_{j+1}$, the set $\langle g\rangle H_j$ is a subgroup of $H_{j+1}$.

And no, you can't infer anything definite about the order of $g_j$ from the information given, although it will be related to the order of $H_j/H_{j-1}$ (do you see how?).

$\endgroup$
2
  • $\begingroup$ Tara B- Thanks for the detailed response. With regard to your last point, i was aiming to say, why is order $r_j=|H_j/H_{j−1}|$ which u mentioned? I am sorry but could you elaborate, sorry for the extra work.. A hunch would be that since the factor group is cyclic and we are adding a term $g_j$ to it, the order of the term $r_j$ would be equal to the order of the factor group which is $|H_j/H_{j−1}|$. Thanks $\endgroup$ – Annonymous Mar 31 '13 at 22:45
  • $\begingroup$ I didn't say that the order of $g_j$ is $|H_j/H_{j-1}$. I don't believe this is necessarily true unless you mean something very specific by "adding an element to a subgroup". $\endgroup$ – Tara B Apr 1 '13 at 22:47

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.