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Let $H \unlhd G$, where $G$ is a finite group, and $ x \in G/H$ of order $p^r$ for some prime $p$ and $r \geq 0$. Then there exists $b \in G$ such that the order of $b$ is $p^r$ and $\pi(b) = x$ where $\pi$ denotes the canoncial projection.

If $r = 0$, we have that the order of $x$ is one which implies that $x$ is the neutral element in $G/H$. Hence $e$ satisfies $\pi(e) = x$ and the order of $e$ is also one.

Thus assume $r > 0$. Now the order of $x$ is $p^r$. Furthermore we can write $x = gH$ for some $g \in G$. Thus we must have $bH = gH$ which implies $b^{-1}g \in H$ or $b = gh$ for some $h \in H$. Furthermore, since the canoncial projection is a homomorphism we have that the order of $gH$ divides the order of $gh$ which implies that the order of $gh$ is $p^s$ for some $s \geq r$. Now by Frobenius we have that $\langle gh \rangle$ has a subgroup of order $p^r$ and since it is a cyclic group also an element of order $p^r$. This element is of the form $(gh)^k$ for some $k \in \mathbb{Z}$. But now I am stucked, somehow it is not possible to show $\pi((gh)^k) = x$ which I think is wrong.

Can somebody help me out?

Edit. With Frobenius I mean:

Suppose the prime power $p^s > 1$ is a divisor of the order of a finite group $G$. Then there is a subgroup of order $p^s$.

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    frobenius?${}{}{}{}$ – 2017-01-06
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    are you sure that you don't have that $H$ is also a finite group, with order coprime to $p$? – 2017-01-06

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As stated the theorem is false, Notice that $\mathbb Z_4$ contains the subgroup $\{0,2\}$. One of its cosets is $\{1,3\}$ and all of its elements have order $4$.

If we also ask that $H$ be finite and $p\nmid |H|$ then it is true.

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    I think the group should be finite, sorry forgot to add. Hmm but the other part I cannot find in the lecture notes. – 2017-01-06
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    doesn't it say somewhere that $p^r$ is the maximum power of $p$ that divides $|G|$ or something? – 2017-01-06
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    Since it is after the chapter on Sylow theorems and in the chapter of the classification of finite abelian groups, this should be the case, I think. – 2017-01-06
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    n that case it is true. – 2017-01-06
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    Thanks. How can I modify my proof in that case? Somehow it does not work that well... – 2017-01-06