Subgroups of finite abelian groups.

Question

For every subgroup $H$ of a finite abelian group $G,$ there exists a subgroup $N$ of $G$ such that $G/N \cong H.$ I need to prove this or give a counter example.

I am aware of isomorphism theorems and classification of abelian groups, direct products etc.

Answer 1

Yes, this is true.

Theorem: Let $G$ be a finite abelian group. The following two statements hold.

(A) Each subgroup of $G$ is isomorphic to a quotient group of $G$,
(B) Each quotient group of $G$ is isomorphic to a subgroup of $G$.

For a proof see [this MSE question]( Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?, which proves $(B)$; but also the same idea works for proving $(A)$. A further reference for the proofs is

L. Fuchs, Abelian Groups. Oxford 1960, page $53$.

Answer 2

the group $G$ is Abelian finite , by the structure theorem $G$ is a direct product of cyclic sub groups $G_i, i=1,...n$ where $|G_1|$ is the exponent of $ G$ and $|G_{i+1}|$ devise $|G_{i}|$ for all $i=1,...,n-1$ also $H$ is a direct product of cyclic sub groups $H_i, i=1,...m$ where $|H_1|$ is the exponent of $ H$ and $m\leq n$, therefor the exponent of $ H$ devise the exponent of $G$. as $G$ is also a direct product of his $p$-Sylow, we get a prove for $G$ p-group and in the end we established the general case.

So we can suppose $n=m$ (because the exponent propriety ensure that we can suppose $H_i$ is in $G_i$ for all $i$), which if necessary to complete by the trivial subgroup. in this case we tack in each $G_i$ a subgroup $N_i$ of order $[G_i:H_i]=$ and form $N$ as the direct product of $N_i$ in $G$, so we obtain $G/N\simeq H$.

in general case is the same argument,decompose $G$ and $H$ in a direct p-Slow, and in each p-Sylow of $G$, tack the appropriate $N_p$ and form $N$ as direct product of his p-Sylow $N_p$

Answer 3

Sketch for a proof of Theorem A [see this answer]

Definition: For an abelian finte $p$-group $G$ define $$ \Omega_i(G) = \{x\in G\mid x^{p^i}=1\}. $$

The following properties hold true:

1. $\Omega_i(G)\subseteq\Omega_{i+1}(G)$.
2. $\Omega_i(G)$ is characteristic.
3. $\Omega_{i+1}(G)/\Omega_i(G) = \Omega_1(G/\Omega_i(G))$.
4. $G=\Omega_1(G)\iff G$ is elementary abelian.
5. If $G=\langle c\rangle$ then $|\Omega_{i+1}(G)/\Omega_i(G)|\le p$, with equality attained if, and only if, $\text{ord}(c)\ge p^{i+1}$.

Part (5) follows from part (3) in the case $G$ cyclic.

Theorem. [Kurtzweil & Stellmacher (2.1.9)]

Let $G$ be a direct product $$ G = A_1 \times \cdots \times A_n \tag{*} $$ of nontrivial cyclic $p$-groups. Define $$ p^{n_i} = |\Omega_i(G)/\Omega_{i-1}(G)|. $$ Then $n_i - n_{i+1}$ is the number of factors of order $p^i$ in $(*)$.

Proof. (Sketch)

For details use the properties above.

Step a. $\Omega_i(G) = \Omega_i(A_1)\times\cdots\times\Omega_i(A_n)$.

Step b. $\Omega_i(A_j) =\Omega_i(G)\cap A_j$.

Step c. $G/\Omega_i(G) \cong A_1/\Omega_i(A_1)\times\cdots\times A_n/\Omega_i(A_n)$.

Step d. \begin{align*} \Omega_{i+1}(G)/\Omega_i(G) &= \Omega_1(G/\Omega_i(G)) &&\text{; prop. 3.}\\ &= \Omega_{i+1}(A_1)/\Omega_i(A_1)\times\cdots\times \Omega_{i+1}(A_n)/\Omega_i(A_n), \end{align*}

Step e. \begin{align*} n_{i+1} &= \{j\mid \Omega_i(A_j)\ne\Omega_{i+1}(A_j)\}\\ &= |\{j\mid|A_j|\ge p^{i+1}\}| &&\text{; prop. 5.} \end{align*} $\Box$

Lemma. $A\leqslant G\implies \Omega_{i+1}(A)/\Omega_i(A)\leqslant\Omega_{i+1}(G)/\Omega_i(G)$.

Proof. The inclusion $A\to G$ induces inclusions $\Omega_j(A)\to\Omega_j(G)$. In particular we have a morphism $$ \Omega_{i+1}(A)/\Omega_i(A)\to\Omega_{i+1}(G)/\Omega_i(G), $$ which is mono because $\Omega_i(A) = \Omega_{i+1}(A)\cap\Omega_i(G)$.

$\Box$

Theorem A [Kurzweil & Stellmacher. Exercise 2.1.7]

If $A$ is a subgroup of $G$, then $A$ is isomorphic to a quotient of $G$.

Proof.

Without loss of generality we may assume that $G$ is a $p$-group. We can further assume that $$ G=(C_{p^{e_1}})^{r_1}\oplus\cdots \oplus(C_{p^{e_i}})^{r_i}\oplus\cdots\oplus(C_{p^{e_n}})^{r_n}, $$ where the $e_1>\cdots>e_n\ge1$ and $r_i>0$ for all $i$. Similarly, we can pick an isomorphism $$ A\cong(C_{p^{d_1}})^{s_1}\oplus\cdots\oplus(C_{p^{d_m}})^{s_m}, $$ with $d_1>\cdots>d_m\ge1$ and $s_j>0$ for all $j$.

From the Lemma and the Theorem above we deduce that there must be $s_1$ factors of $G$ with order $\ge p^{d_1}$, $s_2$ of order $\ge p^{d_2}$ etc. In particular, $d_1\le e_1$ (otherwise, there would be none). More precisely, if $i$ is the maximum index satisfying $d_1\le e_i$, then we must have $r_1+\cdots+r_i\ge s_1$. Thus, we can sum the quotient projections \begin{align*} \varphi_j\colon C_{p^{e_j}}&\to C_{p^{d_1}} \end{align*} to obtain the epimorphism \begin{align*} \varphi=\bigoplus_{j=1}^i\varphi_j\colon (C_{p^{e_1}})^{r'_1}\oplus\cdots \oplus(C_{p^{e_i}})^{r'_i} &\to(C_{p^{d_1}})^{s_1}, \end{align*} where $0\le r'_i\le r_i$ are integers that satisfy $r'_1+\cdots+r'_i=s_1$.

Now, among the remaining factors of $G$, there must be $s_2$ of them with order at least $p^{d_2}$. Thus, we can apply the same argument and repeat it until we exhaust the $m$ summands of $A$. In the end, we will be able to map $s_1+\cdots+s_m$ factors of $G$ onto the decomposition of $A$. The final epimorphism would then consist of this one, plus the mapping of any remaining factors to $0$.

$\Box$