Periodicity is extension-closed

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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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Get more facts about periodic group |Get facts that use property satisfaction of periodic group | Get facts that use property satisfaction of periodic group|Get more facts about extension-closed group property

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.

Related facts

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)