Confusion about the last step of this proof of " Every subgroup of a cyclic group is cyclic":does not subcase $2.2$ contradict the desired conclusion

Source : Reversat & Bigonnet, Algèbre pour la licence ( Undergraduate abstract algebra), Dunod, 1997, p. 20.

Let $G$ be a cyclic group of order $n$ and let $x$ be a generating element of $G$, meaning that $\langle x \rangle = G$.

To be proved : Any subgroup of $G$ is also cyclic.

Note : here, additive notation is used, so every element of $G$ can be written as $kx$ with $0\leq k \leq n-1$.

PROOF [ my question only deals with the last step of the proof, that is with the conclusion ] :

Let $H\subseteq G$ be a subgroup of $G$. Let $k$ be the smallest integer $j$, $1\leq j \leq n-1$ , such that $jx\in H$.

Case 1: $k=1$ If $k=1$ then $H=G$; and since $G$ is cyclic, $H$ is cyclic ( as desired).

Case 2 : k>1

$\space \space \space$ Subcase 2.1 : $k>1$ and there is no $j$, $j\lt k\lt n$, such that $jx\in H$ and $j$ is not a multiple of $k$ ( i.e. $j\ \in \mathbb N$ \ $k\mathbb N$).

In subcase $2.1$, $ H= \{0,kx, 2kx, 3kx... \} = \langle kx\rangle$, and therefore $H$ is cyclic ( as desired) , since $H$ is finite ( being given that $H\subseteq G$).

$\space \space \space$ Subcase $2.2$ $ : k>1$ and there is some $j$, $j\lt k\lt n$, such that $jx\in H$ and $j$ is not a multiple of $k$ ( i.e. $j\ \in \mathbb N$ \ $k\mathbb N$).

In that case, since $j$ ( by hypothsis) is not a multiple of $k$,

• $j = qk+r$ , with $q,r \in \mathbb N$ and $0\lt r \lt k$
• implying that : $jx = qkx + rx$
• implying in turn that : $rx = jx - qkx$.

Since $jx\in H$ ( by hypothesis) and since $qkx\in \langle kx \rangle \subseteq H $ , it follows that $rx \in H$

Subcase $2.2$ yields therefore : if , for some $j$, $j\lt k\lt n$, $j\ \in \mathbb N$ \ $k\mathbb N$, $jx \in H $ then for some $ r \in \mathbb N$ and $0\lt r \lt k$, $rx\in H$.

Conclusion : " Because of the way $k$ has been chosen , one can conclude that any element of $H$ is of the form $pkx$ where $p\in \mathbb N$. In other words: $H = \langle kx\rangle$" .

My question :

I clearly understand how $H = \langle kx\rangle$ follows from Case $1$ and Subcase $2.1$ , but how can it follow from Subcase$2.2$?

I mean, what I understand is that Subcase $2.2$ directly contradicts the conclusion.

As I understand this subcase it means : in case there is only one element of $H$ that is not of the form $pkx$ ( since $j$, by assumption is not a multiple of $k$ ) there is another element $rx$ that is not of the form $pkx$ ( since , by constructon, $rx= (j-qk)x$, implying that $r$ cannot be written as $pk$).

Or am I wrong as to my interpretation of the conditions imposed on $j$?

Solution 1:

The last step is a proof by contradiction. $k$ was chosen to be the smallest, but in step 2.2 you've found $r$ with the same property but smaller than $k$.