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”Gを群とし、A,BをGの巡回部分群とする。 G=ABを満たし、A,Bが共にGの正規部分群とならない例を挙げよ。”

”A,Bが共にGの正規部分群とならない”は、二通りに解釈できて
・「A,Bが共にGの正規部分群」とならない と解すれば、半直積の場合が適合する
・「A,Bが共にGの正規部分群とならない」＝A,B両方が非正規部分群 と解すれば、半直積の場合は不適
積 G=AB は、半直積以上を考える必要がある

和語では、良い情報がヒットしないので
英語で検索したら、下記ヒットした
”Zappa–Szép product”使えだってさ （＾＾
wikipediaのReferences に(2007)の文献とか上がっている
なお当然だが、AIは100％は信用できない。各人検証乞う

(google検索)
In a group G, are there two subgroups A and B that satisfy the group product G=AB, where neither A nor B are normal subgroups of G?
AI による概要
Yes, it is possible for a group \(G\) to be the product of two subgroups \(A\) and \(B\) (written as \(G=AB\)) where neither \(A\) nor \(B\) is a normal subgroup of \(G\).
This type of group factorization is known as a Zappa–Szép product (or a general product). It generalizes the more common semi-direct product, where at least one factor is required to be normal.
Concrete Example: The Symmetric Group \(S_{4}\)
The smallest non-trivial example of this phenomenon occurs in the Symmetric Group \(S_{4}\) (which consists of all 24 permutations of the set \(\{1, 2, 3, 4\}\)).
1.Group \(G\): Let \(G = S_4\) (order 24).
2.Subgroup \(A\): Let \(A\) be the subgroup of all permutations that fix the number 4. This subgroup is isomorphic to \(S_{3}\) and has an order of 6.
・Normality Check: \(A\) is not normal in \(S_{4}\). For instance, conjugating a permutation in \(A\) by an element that moves "4" (like the transposition \((14)\)) will result in a permutation that fixes "1" instead of "4."
3.Subgroup \(B\): Let \(B\) be a cyclic subgroup of order 4 generated by a 4-cycle, such as \(B = \langle (1234) \rangle = \{e, (1234), (13)(24), (1432)\}\).
・Normality Check: \(B\) is not normal in \(S_{4}\). In \(S_{4}\), the only proper normal subgroups are the Alternating Group \(A_{4}\) and the Klein four-group \(V_{4}\). \(B\) is not either of these.

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