Subgroup Generated by Subgroup and Element

== Theorem ==
Let $G$ be a finite abelian group.
Let $H$ be a proper subgroup of $G$.
Let $a \in G \setminus H$.
Let $n$ be the indicator of $a$ in $H$.
Then:
:$K = \left\{{x a^k: x \in H, \ 0 \le k < n}\right\}$
is a subgroup of $G$ such that $H \subseteq K$, and each element of $K$ has a unique representation in this form.
Moreover:
:$K = \left\langle{H, a}\right\rangle$
where $\left\langle{H, a}\right\rangle$ denotes the subgroup generated by $\left\{{H, a}\right\}$, and:
:$\left\vert{K}\right\vert = n \left\vert{H}\right\vert$
where $\left\vert{K}\right\vert$ denotes the order of $K$.
== Proof ==
=== $K$ is Subgroup of $G$ ===
We first show that $K$ is a subgroup of $G$ using the Two-Step Subgroup Test.
$(1): \quad K \ne \varnothing$:
From Identity of Subgroup:
:$e \in H$
From Indicator is Well-Defined, $n > 0$ and so $n - 1 \ge 0$.
Thus $k = 0$ fulfils the condition that $0 \le k < n$, and so:
:$e = e a^0 \in K$
Thus $K \ne \varnothing$.
$(2): \quad K$ is closed:
Let $r = x a^k \in K$ and $s = y a^l \in K$ where $x, y \in H$ and $0 \le k < n, 0 \le l < n$.
$G$ is abelian, so $r$, $s$ and $a$ commute.
So:
:$r s = x y a^{k + l}$.
Since $H$ is a group and $x, y \in H$ then also $x y \in H$.
Let $k + l \le n - 1$.
Then $r s \in K$ by definition of $K$.
Let $n \le k + l \le 2 n$.
Then:
:$r s = x y a^n a^{k + l - n}$
By definition of indicator, $a^n \in H$.
Thus as $x, y, a^n \in H$ it follows that $x y a^n \in H$.
As $0 \le k + l - n \le n - 1$ it follows that $\left({x y a^n}\right) a^{k + l - n} \in K$ by definition of $K$. 
So again $r s \in K$.
Thus $K$ is closed.
$(3): \quad K$ is closed under inversion:
Let $r = x a^k \in K$ where $x \in H$ and $0 \le k < n$.
Then $x^{-1} \in H$.
Let $k = 0$.
Then 
:$r^{-1} = x^{-1} e = x^{-1} a^0 \in K$
Let $k > 0$.
Then:
:$r^{-1} = x^{-1} a^{-k} = x^{-1} a^{-n} a^{n - k}$
By definition of indicator, $a^n \in H$.
We have that $x^{-1}, a^n \in H$.
As $G$ is abelian, $x$ and $a$ commute.
Thus from Inverse of Commuting Pair:
: $x^{-1} a^{-n} = \left({x a^n}\right)^{-1} \in H$
Also:
:$0 \le n - k < n -1$
so:
:$r^{-1} = \left({x a^n}\right)^{-1} a^{n - k} \in K$
Therefore by the Two-Step Subgroup Test we have shown that $K$ is a subgroup of $G$.
=== $H$ is Subgroup of $K$ ===
As:
:$\forall x \in H: x = x a^0 \in K$
and so $H \subseteq K$.
=== Order of $K$ ===
It remains to be shown that:
:$\left\vert{K}\right\vert = n q$
where $q = \left\vert{H}\right\vert$.
Let $x a^k = y a^l \in K$.
WLOG, let $k \ge l$ (if not, just relabel the two).
We have:
: $a^{k - l} = x^{-1} y$
Since $H$ is a subgroup of $G$:
:$x^{-1} y \in H$
and therefore:
:$a^{k - l} \in H$
By definition of indicator, $n$ is the least (strictly) positive integer such that $a^n \in H$
Hence because $n > k \ge k - l$:
:$k - l = 0$
Therefore $k = l$, and $x = y$.
Therefore the elements $x a^k$ with $x \in H$ and $0 \le k < n$ are distinct.
So for each $x \in H$ there are $n$ elements of the form $x a^k$ in $K$.
We have that there are $q$ elements in $H$.
Thus it follows that there are $n q$ such elements.
By definition of $K$, these form all of $K$.
Thus:
:$\left\vert{K}\right\vert = n q$
as required
Category:Subgroups