# Characterization of groups of order 110 (understanding semidirect product)

I try to characterize all groups $$G$$ of order $$110=2 \cdot 5 \cdot 11$$.

Sylow implies the existence of a subgroup $$N \trianglelefteq G$$ of order $$55$$ and a subgroup $$H$$ of order $$2$$. It follows, that $$G$$ is a semidirect product $$N \rtimes H$$.
As there exists only two groups of order $$55$$, namely $$C_{55}$$ and $$C_{11}\rtimes C_5$$, it follows that $$G$$ is of the form $$G_1=C_{55} \rtimes C_2$$ or $$G_2=(C_{11}\rtimes C_5) \rtimes C_2$$.
Via GAP I found that there exist $$4$$ groups of the form $$G_1$$. I was wondering how this is possible. To my understanding, the semidirect product ist defined by an homomorphism $$\phi : C_2 \mapsto Aut(C_{55})=(C_{55})^\times=C_{18}$$. Of course there is the trivial homomorphism leading to the group $$C_{110}$$. Am I right, that there exists only one other homomorphism, sending $$1$$ to $$9$$, which leads to $$D_{110}$$?
How is it possible, that there are two more semidirect products of this kind, abstractly $$C_5 \times D_{22}$$ and $$C_{11} \times D_{10}$$, both with normal $$C_{55}$$-Subgroup. To which mapping do belong?

Furthermore I would like to know how it can be seen, that there exist only two groups of type $$G_2$$, namely $$(C_{11}\rtimes C_5) \times C_2$$ and $$(C_{11}\rtimes C_5) \rtimes C_2 \cong C_{11} \rtimes C_{10}$$. Do I need to calculate the autmorphism group of $$C_{11}\rtimes C_5$$?

You've got it wrong; $$\mathbb Z_{55}^×$$ has order $$\varphi(55)=40$$ and is isomorphic to $$\mathbb Z_4×\mathbb Z_{10}$$. The three non-direct semidirect products are derived from the three elements of order $$2$$ in $$\mathbb Z_{55}^×$$: $$34$$, $$21$$ and $$-1$$.

• Ahhhh thanks, this tiny mistake made me mad. Jan 4, 2021 at 12:00

In "Die Gruppen der Ordnungen p^3,pq^2,pqr,p^4", O.Hoelder, Math. Annalen, XLIII, 1894

you can find all necessary informations.

Shortly, the are six groups (GAP lists all these groups). They are

1. Z110
2. D110
3. Z5 x D22
4. Z11 x D10
5. Z2 x (Z11 semidirect Z5)
6. Z11 semidirect Z10

D. Kaesbauer