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I am struggling to understand the proof of the following theorem:

Let G' be a subgroup of a finite abelian group G, where G' $\neq$ G. Choose an element a in G, a $\notin$ G', and let h be the indicator of a in G'. Then the set of products

$$ G'' = {xa^k:x \in G' and k = 0,1,2,....., h-1} $$ is a subgroup of G which contains G'. Moreoever, the order of G'' is h times that of G', $$ \mid G'' \mid = h\mid G'\mid $$

The proof runs as follows:

First we test closure. Choose two elements in G'', say $xa^k$ and $ya^j$, where x,y $\in$ G' and $0 \leq k < h, 0 \leq j < h$. Since G is abelian the product of the elements is $$

\begin{equation} (xy)a^{k+j} \tag{1} \end{equation} Now $k +j = qh + r \tag{2}$ where $0 \leq r < h$

I understand the rest of the proof if this claim is granted, but I'm not sure I follow the resoning here.

We know that k, j are both less than h, but why does it necessarily follow that k+j are greater than h, which would in fact imply that $k+j = qh + r$ (by Euclid's Algorithm).

I feel like I'm missing a trivial trick here....

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    What does "Let G'' be a subgroup of a finite abelian group G, where G' $\neq G\;$ " mean? What's the relation between $\;G,\,G',\,G''\;$ in that first line? What's the *indicator* of an element in a group? – 2017-02-12
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    My apologies, finger must have slipped. The first group is `G'` – 2017-02-12
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    I'm not sure I understand the question..... if $G' \leq G$ it follows that there may be elements $a \in G \notin G'$. If $G' < G$ it follows that there *will* be elements $a \in G \notin G'$... or am I misunderstanding something? – 2017-02-12

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If I understand correctly, then $h$ is the order of $a$ in $G$, which is to say the least number such that $a^h = 1$

One way to think about $G''$ here is that it is the union of cosets $x\langle a \rangle$, where $x$ in $G'$ and $\langle a \rangle$ is the cyclic group generated by $a$. The argument that $G''$ is closed can be concisely summarized by saying that $G$ is abelian and both $\langle a \rangle$ and $G'$ are closed.

For the part in question, we're just saying that, since $h$ is the order of $a$. $$a^ja^k = a^{j+k} = a^{qh+r}=a^{qh}a^r=a^r$$ It's ok if $k+j < h$, in that case $q=0$ and $r=k+j$.

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    derp... that last comment was exactly what I was missing. Thanks! – 2017-02-12