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

$\blacksquare$