Subgroup Generated by Subgroup and Element

Theorem
Let $G$ be a finite abelian group, and $H$ a proper subgroup of $G$.

Let $a \in G \backslash H$, and $n$ the indicator of $a$ in $G$.

Then:


 * $K = \{ x a^k : x \in H,\ 0 \leq k \leq n - 1\}$

is a subgroup of $G$ containing $H$, and each element has a unique representation in this form.

Moreover, $K = \langle H, a\rangle$, and $|K| = n|H|$, where $|\cdot |$ denotes the order of a group.

Proof
We first show that $K$ is a subgroup using the Two-Step Subgroup Test.

1. $K \neq \emptyset$'':

This is true because $H \neq \emptyset$ and $n - 1 \geq 0$, $K \neq \emptyset$.

Therefore $e = e a^0 \in K$.

2. $K$ is closed under multiplication:

Suppose $r = x a^k$ and $s = y a^\ell$ lie in $K$.

We have $rs = xy a^{k+\ell}$. Since $H$ is a group and $x,y \in H$ then also $xy \in H$.

If $k + \ell \leq n - 1$, then $rs \in K$.

If $n \leq k + \ell \leq 2n$, then $rs = xya^n a^{k+\ell - n}$, with $xya^n \in H$ and $0 \leq k+\ell - n \leq n-1$.

So again $rs \in K$, and $K$ is closed under multiplication.

3. $K$ is closed under inversion:

Let $r = x a^k$ as above. If $k = 0$, then


 * $r^{-1} = x^{-1} \in K$.

If $k > 0$ we have


 * $r^{-1} = x^{-1}a^{-k} = x^{-1}a^{-h} a^{h-k}$

Since $x, a^h \in H$, also $x^{-1}a^{-h} = (xa^h)^{-1} \in H$.

Moreover $0 \leq h - k < n -1$, so $r^{-1} \in K$.

Therefore by the Two-Step Subgroup Test we have shown that $K$ is a subgroup of $G$.

Clearly $H \subseteq K$, since $x = x a^0 \in K$ for all $x \in H$.

It remains to show that $|K| = n q$, where $q = |H|$.

Let $x a^k = y a^\ell \in K$. Without loss of generality, we let $k \geq \ell$ (if not, just relabel the two).

We have $a^{k -\ell} = x^{-1}y$. Since $H$ is a subgroup of $G$, $x^{-1}y \in H$, and therefore $a^{k-\ell} \in H$.

Because $n > k \geq k-\ell$, and $n$ is the least positive integer such that $a^n \in H$, we must have $k-\ell = 0$.

Therefore $k = \ell$, and $x = y$.

Therefore the elements $x a^k$ with $x \in H$ $0 \leq k \leq n-1$ are distinct.

Evidently there are $nq$ such elements, and by definition these form all of $K$.

Thus $|K| = nq$ as required.