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Let $G$ be a group and $A$ and $B$ be two normal subgroups of $G$.

Suppose $f$ is a homomorphism from $G$ to $B$ with kernel equal to $A$ and image equal to $B$. By the first isomorphism theorem, we have $G/A \cong B$. Thinking extremely naively, we might hope to be able to "re-arrange" this equation and guess that $G/B \cong A$. Of course, group quotients don't always behave like numbers, so:

  1. Is $G/B \cong A$? Or in other words, does there always exist another homomorphism $g:G \to A$ such that the kernel of $g$ is equal to $B$ and the image of $g$ is equal to $A$?

  2. If yes, then how can we construct such a $g$?

If the answer to question (1) is in general no, then are there any specific conditions under which such a homomorphism exists and can be constructed?

Noah Caplinger
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Ravi
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    @MartinBrandenburg I don't see how this is an XY problem. OP just offered an alternative (but equivalent) formulation of the title of the post. – Charles Hudgins Dec 23 '24 at 15:56

1 Answers1

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The general answer to question 1 is 'no', but a slight modification of the hypotheses makes it a 'yes'. This modified question and answer sheds some light on the original question.

Modified question: Let $A,B \subset G$ be normal, and suppose there is a homomorphism $G \to B$ with kernel $A$ and whose restriction to $B$ is an isomorphism. Then is there a corresponding map $G \to A$ with kernel $B$?

Answer: yes, and in fact $G \cong A \times B$.

Proof sketch: Since the restriction of $f$ to $B$ is an isomorphism, and $ker(f)=A$, the two subgroups $A$ and $B$ are disjoint and generate $G$. Then for all $a\in A$, $b \in B$, we know that $$aba^{-1}b^{-1} = (aba^{-1})b^{-1} = a(ba^{-1}b^{-1})$$ is in both $A$ and $B$, since they are normal. Then $aba^{-1}b^{-1}$ is trivial and hence all elements of $A$ and $B$ commute. Since $A$ and $B$ together generate $G$, we conclude $G \cong A \times B$.

Alternate, (harder) but more enlightening proof sketch: The map $f$ splits, so $G$ decomposes as a semidirect product $G = A \rtimes B$ (see here). Since $B$ is normal, the action of $B$ on $A$ must be trivial. A semidirect product with a trivial action is just a direct product.

The upshot is that the condition $G/A \cong B$ occasionally allows us to decompose the ambient group into pieces related to $A$ and $B$. The most powerful way to make this happen is the "splitting criterion" which is essentially the condition I added earlier.

In order to find a counterexample to the original statement, we need to look for $A,B \subset G$ so that $G/A\cong B$, but the quotient map is not an isomorphism when restricted to $B$. In the jargon of group extensions, we are searching for a non-split extension. We also have the additional requirement that $A$ and $B$ be normal. The klein-bottle group $$G = \langle a,b \mid bab^{-1} = a^{-1}\rangle$$ works: one can check that $A = \langle a\rangle$ and $B = \langle b^2\rangle \cong \mathbb{Z}$ are normal, and that $G/A \cong \mathbb{Z} \cong B$, but the image of $B$ under the quotient map is $2\mathbb{Z} \subset \mathbb{Z}$. However, $G/B \cong D_\infty\not \cong A$, so this works as a counterexample.

Noah Caplinger
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  • Very nice and complete answer. Another example for this "failing" that might be more familiar to newcomers to group theory is $C_4 \times C_2$ and its non-cyclic quotient. – Tobias Kildetoft Dec 27 '24 at 11:03