Periodicity is extension-closed

This article gives the statement, and possibly proof, of a group property (i.e., periodic group) satisfying a group metaproperty (i.e., extension-closed group property)
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Statement

Suppose $G$ is a group and $H$ is a normal subgroup of $G$ such that both $H$ and the quotient group $G/H$ are periodic groups. Then, $G$ is also a periodic group.

Further, if $\pi$ is a set of primes such that both $H$ and $G/H$ are $\pi$-groups -- all prime factors of orders of elements in these groups are in $\pi$ -- then $G$ is also a $\pi$-group.

Proof

We prove the second version. Note that taking $\pi$ to be the set of all primes will give the first version.

Given: A group $G$. A normal subgroup $H$ with quotient group $G/H$. A prime set $\pi$ such that both $H$ and $G/H$ are $\pi$-groups: all elements of $H$ and all elements of $G/H$ have finite orders with all prime divisors of the order of every element in $\pi$.

To prove: Every element of $G$ has finite order and every prime divisor of the order of every element is in $\pi$.

Proof: We pick an arbitrary element $g \in G$. We let $\varphi:G \to G/H$ be the quotient map.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $\varphi(g)$ has order equal to some $\pi$-number $m$ as an element of $G/H$. $G/H$ is a $\pi$-group.
2 $g^m \in H$ where $m$ is the $\pi$-number chosen in Step (1). Step (1)
3 Let $h = g^m$ from Step (2). Then, $h$ has order equal to some $\pi$-number $n$ as an element of $H$. $H$ is a $\pi$-group Step (2)
4 $(g^m)^n$ is the identity element of $G$. Thus, the order of $g$ in $G$ is finite and divides the $\pi$-number $mn$, hence it must also be a $\pi$-number. Step (3)