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Let $A$ be an Abelian group and suppose $A$ has a subgroup isomorphic to $\mathbb{Z}/p^a\mathbb{Z}$, for some prime $p$ and positive integer $a$. Suppose that $A/(\mathbb{Z}/p^a\mathbb{Z}) \cong \mathbb{Z}/p^b\mathbb{Z}$, for some positive integer $b$.

How can I show that $A$ is a finite group? I added these extra hypotheses for my question, but in general, if $A$ is some group and $B \leq A$ is a finite normal subgroup of $A$ and $A/B \cong C$ for some finite group $C$, does this imply that $A$ is finite?

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    $\begingroup$ Certainly, with $|A|=|B||C|$. $\endgroup$
    – the_fox
    Feb 1, 2019 at 1:56
  • $\begingroup$ But that assumes that $A$ is a finite group. This is what we want to show. $\endgroup$ Feb 1, 2019 at 1:56
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    $\begingroup$ @FredericChopin No it doesn't, that formula is valid regardless of the cardinalities of the groups involved. $\endgroup$ Feb 1, 2019 at 2:23

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Let $A$ be our general group, and let $B$ be a finite, normal subgroup of $A$. Let’s say that $|B| = k$. Then the following things are true:

  • For any $x\in A$, the coset $xB$ satisfies $|xB| = |B| = k$.

  • If $x,y\in A$, then $xA\cap yA = \emptyset$ if and only if $x\notin yA$. That is, the cosets form a partition of $A$.

  • The size of $A/B$ is (by definition) the number of distinct cosets of $B$ in $A$.

If $A/B$ has $m$ elements, that means that $A$ can be partitioned into $m$ disjoint sets, each of which has $k$ elements. Thus $|A| = mk$.

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  • $\begingroup$ These were very helpful hints, thank you. I was able to prove the facts that you claimed were true. $\endgroup$ Feb 1, 2019 at 3:13
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The "slick" way is Lagrange's theorem, which says tells you that the number of elements of $A$ is equal to the number of elements of $B$ times the number of elements of $A/B$. This statement has a rigorous formulation independent of whether the groups are finite or infinite, and it implies that if two of three sets $A, B, A/B$ are finite, then so is the third.

Another way (basically the same thing): let $n$ be the number of elements of $B$. Take an element $a \in A$, and consider its image $\overline{a}$ in the finite group $A/B = C$. Show that there are exactly $n-1$ other elements of $A$ having the same image. Since there are only finitely many possibilities for $\overline{a}$, this shows that there are only finitely many elements of $A$.

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Yes. Using the first isomorphism theorem, $\mid C\mid=\frac{\mid A\mid}{\mid B\mid}$. Just consider the quotient map.

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