# Normal subgroup contained in the hypercenter satisfies the subgroup-to-quotient powering-invariance implication

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., normal subgroup contained in the hypercenter) must also satisfy the second subgroup property (i.e., normal subgroup satisfying the subgroup-to-quotient powering-invariance implication)
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## Statement

### Original formulation

Suppose $G$ is a group and $H$ is a normal subgroup contained in the hypercenter of $G$. Suppose $p$ is a prime number such that $G$ and $H$ are both powered over the prime $p$. Then, the quotient group $G/H$ is also powered over $p$.

### Corollary formulation

Suppose $G$ is a group and $H$ is a normal subgroup contained in the hypercenter of $G$ that is also a powering-invariant subgroup of $G$. Then, $H$ is a quotient-powering-invariant subgroup of $G$.

## Facts used

1. Upper central series members are normal (in fact, they are strictly characteristic subgroups)
2. Normality is strongly intersection-closed
3. Upper central series members are powering-invariant
4. Central implies normal satisfying the subgroup-to-quotient powering-invariance implication
5. Third isomorphism theorem

## Proof

### Case of containment in a member of the finite upper central series

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Note that this case suffices for finite groups and nilpotent groups.

Given: A group $G$ with upper central series $Z^0(G) \le Z^1(G) \le \dots$. A normal subgroup $H$ of $G$ contained in $Z^n(G)$ for some positive integer $n$. A prime number $p$ such that both $G$ and $H$ are powered over $p$.

To prove: $G/H$ is powered over $p$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the subgroup series $1 = H \cap Z^0(G) \le H \cap Z^1(G) \le H \cap Z^2(G) \le \dots \le H \cap Z^n(G) = H$. For $i \ge n$, $H \cap Z^i(G)$ is stable at $H$. $H$ is contained in some $Z^n(G)$.
2 Each subgroup of the form $H \cap Z^i(G)$, $i$ a nonnegative integer, is a normal subgroup of $G$. Facts (1), (2) $H$ is normal in $G$. Fact-given direct.
3 For any nonnegative integer $i$, $(H \cap Z^{i+1}(G))/(H \cap Z^i(G))$ is central in $G/(H \cap Z^i(G))$. $H$ is normal in $G$. Step (2) (for making sense of quotients) We have $[G,H \cap Z^{i+1}(G)] \le [G,H] \cap [G,Z^{i+1}(G)] \le H \cap Z^i(G)$, and the claim follows.
4 For any nonnegative integer $i$, we have that if $G/(H \cap Z^i(G))$ is $p$-powered and $(H \cap Z^{i+1}(G))/(H \cap Z^i(G))$ is $p$-divisible (here, $p$-divisible means every element has a not necessarily unique $p^{th}$ root in the group), then $G/(H \cap Z^{i+1}(G))$ is $p$-powered. Facts (4), (5) Step (3) First note that if $G/(H \cap Z^i(G))$ is $p$-powered, then any $p$-divisible subgroup must be $p$-powered because of global uniqueness of $p^{th}$ roots. Thus, $(H \cap Z^{i+1}(G))/(H \cap Z^i(G))$ is $p$-powered. Step (3) tells us that $(H \cap Z^{i+1}(G))/(H \cap Z^i(G))$ is central in $G/(H \cap Z^i(G))$, hence by Fact (4) and the preceding sentences, we get that the quotient group $\frac{G/(H \cap Z^i(G))}{H \cap Z^{i+1}(G))/(H \cap Z^i(G))}$ is $p$-powered. By Fact (5), this is isomorphic to $G/(H \cap Z^{i+1}(G))$, completing the proof.
5 Each of the subgroups $H \cap Z^i(G)$ is $p$-powered and hence $p$-divisible. Fact (3) $G$ and $H$ are both $p$-powered. By Fact (3), each $Z^i(G)$ is $p$-powered. Combine with the fact that $H$ is $p$-powered to get that the intersection is.
6 Each quotient group $(H \cap Z^{i+1}(G))/(H \cap Z^i(G))$ is $p$-divisible (here, $p$-divisible means every element has a not necessarily unique $p^{th}$ root in the group). Step (5) This follows directly from Step (5), and the observation that any quotient of a $p$-divisible group by a normal subgroup is $p$-divisible.
7 For any nonnegative integer $i$, we have that if $G/(H \cap Z^i(G))$ is $p$-powered, so is $G/(H \cap Z^{i+1}(G))$. Steps (4), (6) Step-combination direct.
8 $G/(H \cap Z^i(G))$ is $p$-powered for all $i$. $G$ is $p$-powered. Step (7) Step (7) and use the principle of mathematical induction, with base case $i = 0$ (i.e., $G$ is $p$-powered).
9 $G/H$ is $p$-powered. $H \le Z^n(G)$ for some $n$. Step (8); instead of given, can use Step (1) Set $i = n$ in Step (8).

### Modification for transfinite upper central series case

We need to modify the above to do transfinite induction. The main change is that the $i$ will everywhere be interpreted as an ordinal (possibly infinite). In addition, Step (7) will now need an accompanying step that handles the limit ordinal situation. The handling of the limit ordinal situation is similar to the proof for quotient-powering-invariance is union-closed, but done in slightly greater generality.