Order element of a quotient group

Question

Let $G$ be a finite group, $H$ a normal soubgrup of $G$ and $|G/H|=n$. Let a $\in G$. If $|a| = m$ then $|aH|=(m,n)$.

Because of $a^{n}H = H$ and $a^{m}H = H$, then $|aH||(m,n)$. Suppose that $|aH| < (m,n)$:

$a^{(m,n)-|aH|}H = H$

This means that $2(m,n) = |aH|$ or $(m,n)=|aH|$.

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Is this razoning right?. If so, how can I end this proof?

Answer 1

This is false. Let $G= \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, and let $H$ be the second factor. Then $|G/H|=2$. Let $a=(0,1) \in H$. Then $|a|=2$ so $(|a|,|G/H|)=2$, but $|aH|=1$ since $a \in H$.

Answer 2

This is unclear and wrong. If $aH$ indicates a coset, it has the same size as $H$. If $aH$ indicates the subgroup generated by $a$ and $H$, which should be denoted $<a,H>$, the group of the square gives a counterexample. Let $H$ be the 4 rotations and $a$ be any reflection. The. $H$ has order 4, $a$ order 2, but they generate the entire group, which has order 8.

The formula you want is that the size of the set of pairwise products of two subgroups (which is it itself a subgroup of either of them is normal) is the product of the sizes divided by the size of the intersection.