Finite index implies powering-invariant

Statement
Suppose $$G$$ is a group and $$H$$ is a subgroup of finite index in $$G$$. Then, $$H$$ is a powering-invariant subgroup of $$G$$. In other words, if $$p$$ is a prime number such that $$G$$ is powered over $$p$$, then $$H$$ is also powered over $$p$$.

Facts used

 * 1) uses::Poincare's theorem states that any subgroup of finite index contains a normal subgroup of finite index
 * 2) uses::Normal of finite index implies quotient-powering-invariant
 * 3) uses::Finite implies powering-invariant
 * 4) uses::Powering-invariant over quotient-powering-invariant implies powering-invariant

Proof
Given: A group $$G$$, a subgroup $$H$$ of finite index in $$G$$.

To prove: $$H$$ is powering-invariant in $$G$$.

Proof: