Quotient Group Is Finite Implies That Group Is Finite?

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?

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

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 $$|A| = mk$$.

• These were very helpful hints, thank you. I was able to prove the facts that you claimed were true. Feb 1, 2019 at 3:13

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 $$A$$.

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