# Abelian Group Extension

I have the following group:

$$A = \mathbb{Z}/p_1^{r_1} \times \mathbb{Z}/p_2^{r_2} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_n^{r_n} \mathbb{Z}$$

each $p_i$ being a prime, but they need not be distinct primes.

I want to solve the following isomorphism of abelian groups,

$$\frac{X}{\mathbb{Z}/p\mathbb{Z}} \cong A$$

I think there are only two types of possibility:

• Direct product $X = \mathbb{Z}/p\mathbb{Z} \times A$
• If $p = p_i$ we could have $\mathbb{Z}/p_1^{r_1} \times \mathbb{Z}/p_2^{r_2} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_i^{r_n+1} \mathbb{Z} \times \cdots \times \mathbb{Z}/p_n^{r_n} \mathbb{Z}$

But how could I prove these are the only possibility? thanks a lot!

Suppose $X$ is such a group. Consider the collection of all elements of order $p$ in $X$; it is not empty, since $p$ divides the order of $X$, so by Cauchy's Theorem there is such an element. If there exists $x\in X$ of order $p$ such that $\langle x\rangle \cap A = \{0\}$, then let $B=\langle x\rangle$. Then $A\cap B=\{0\}$, and $|A+B| = |X|$, hence $A+B=X$, and therefore $X=A\oplus B$, proving that $X\cong A \times C_p$, as desired (I use $C_n$ for the cyclic group of order $n$).
The only other possibility is that every element of order $p$ in $X$ is contained in $A$. Let $x\in X$ be an element of order a power of $p$ that maps to a generator of $X/A$ (it exists, because if you pick any preimage of one generator, and ther order is not a $p$-power, you can pick an adequate multiple to get another element that maps to perhaps a different generator and is a prime power. Note also that $x\notin A$ and $px\in A$. So now we are working inside the $p$-parts of $X$ and $A$, and so we may assume that $X$ and $A$ are both abelian $p$-groups. So $A=C_{p^{a_1}}\oplus\cdots\oplus C_{p^{a_k}}$ with $a_1\leq\cdots\leq a_k$. Write $px$ as an element of $A$; by adding suitable elements of $A$ to $x$ you can ensure that any component of $px$ that is a multiple of $p$ is actually equal to $0$: for example, if the $i$th component is $p^{\ell}ky_i$, where $y_i$ is the generator, $\ell\gt 0$ and $\gcd(k,p)=1$, then adding $-p^{\ell-1}ky_i$ to $x$ still keeps it an element that maps to a generator of $X/A$, but now the $i$th component of $px$ is $0$. Then replacing any generators by suitable powers, you may assume that the "vector" that describes $px$ has all components equal to either $0$ or $1$. Doing an easy change of basis for $A$ gives you the description of $X$ that you want in this situation.
The abelian group $X$ has order equal to $p \times$ the order of $A$, and contains a subgroup of order $p$ such that the quotient is isomorphic to $A$. If you think about the possible elementary divisors of $X$ which are compatible with these two statements, you will get the result you want.