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 $g_j$?

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    $\begingroup$ 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

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?).

  • $\begingroup$ 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

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