How can this subgroup $H$ of a cyclic group $G=\langle x \rangle$ contain the identity element if $H= \{1x, 2x, 3x, .... \} $? 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 $0\leq k \leq n-1$.
The proof I find in the above mentionned book  begins like this ( I only quote the beginning, this my question bears on this part of the proof) :

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

My problem  :
I understand that the assumption according to which $H$ is a subgroup of $G$ implies by itself that $0$ ( the identty element) belongs to $H$.
But on the other side, the hypothesis " $k$ is the smallest $j$ $1\leq j \leq n-1$ ,  ..." seems to imply that :
$H= \{1x, 2x, 3x ... \}$,
and I cannot see how, under these conditions, $H$ could contain $0x=0$.
What do I miss?
 A: Definition of ''multiple'' of a group element $x\in G$:
$0x=0$ and $(k+1)x = kx + x$ for all $k\geq 0$.
Then in particular, $1x = 0x +x = 0+x=x$.
A: 
But on the other side, the hypothesis "$k$
is the smallest $j$ $1\leqslant j\leqslant n-1$..." seems to imply that: $H = \{1x,2x,\ldots,\}$.

Consider the set of integers $A = \{-2,-1,0,3,4,7\}$. For  $n=7\in\mathbb N$, what is the smallest $j$ with $1\leqslant j\leqslant 6$ that is in $\mathbb Z$? It is $3$. Does this imply that $0\notin \mathbb Z$? No. But $j=0$ does not satisfy the additional condition that $1\leqslant  j\leqslant 6$.
In short, you are taking the set $A = \{ j\in \mathbb Z : jx\in H\}$ and taking its intersection with $B = \{1,2,\ldots,n-1\}$ and looking at $\min(A\cap B)$. Of course $A\cap B$ misses some elements of $A$, but this is not going to change what $H$ looks like.
For example, if $n=6$ so that $G = \mathbb Z/6$, you can consider $H = \{0,2,4\} =\langle 2 \rangle$ the subgroup generated by $2$, isomorphic to $\mathbb Z/3$. In this case
$$A = \{\ldots,-4,-2,0,2,4,\ldots\}$$
consists of all even integers (positive and negative) and $A\cap B = \{2,4\}$ so $\min(A\cap B) =2$.
