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    $A/A[p]$ versus $A^p$, and $A/A^p \cong A[p]$ when $A$ is finite

    P-A2PAP

    Area
    Algebra
    Topics
    Abelian Groups, Groups
    Status
    draft

    Let \(A\) be an abelian group (we will write the group multiplicatively in this problem). For a prime number \(p\), define the following subgroups of \(A\): \[A^p := \{a^p : a \in A\}, \qquad A[p] := \{a \in A : a^p = 1\}.\]

    1. Prove or give a counterexample: \(A/A[p] \cong A^p\).

    2. Prove: If \(A\) is finite, then \(A/A^p \cong A[p]\).

    Solution

    1. The assertion in part (a) is true for every abelian group \(A\).

    The power map \(u:A\to A\), given by \(u(a)=a^p\), is a homomorphism because \(A\) is abelian: \((ab)^p=a^pb^p\). Its kernel is \(A[p]\) and its image is \(A^p\), so these are indeed subgroups. Define \[ \overline u:A/A[p]\longrightarrow A^p, \qquad aA[p]\longmapsto a^p. \] It is well-defined: changing \(a\) to \(ac\) with \(c^p=1\) does not change \(a^p\). It is a homomorphism and is onto by the definition of \(A^p\). If \(a^p=b^p\), then \((b^{-1}a)^p=1\), so \(aA[p]=bA[p]\); thus it is also injective, and \(A/A[p]\cong A^p\).

    2. If \(A\) is finite, then \(A/A^p\) and \(A[p]\) have equal orders.

    Each fiber of \(u\) is a coset of \(A[p]\), so counting fibers gives \(|A|=|A[p]|\,|A^p|\). Therefore \[ |A/A^p|=\frac{|A|}{|A^p|}=|A[p]|. \]

    3. Under the same finiteness assumption, the groups in part (b)are isomorphic.

    Both groups are abelian and have exponent dividing \(p\). For \(A[p]\) this is its defining property. For the quotient, \[ (aA^p)^p=a^pA^p=A^p, \] since \(a^p\in A^p\).

    An abelian group \(B\) of exponent dividing \(p\) is a vector space over \(\mathbb F_p\): vector addition is the group operation and scalar multiplication by the residue class of \(n\) is \(b\mapsto b^n\). This is independent of the representative \(n\) because \(b^p=1\); the vector-space identities follow from the laws of exponents and commutativity. The group identity is the zero vector.

    2. A finite-dimensional \(\mathbb F_p\)-spaceof dimension \(r\) has \(p^r\) elements, so their dimensions are equal.Choosing a basis in each and matching the bases gives a linearisomorphism and hence the asserted group isomorphism. This alsocovers the trivial groups, with dimension zero.

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