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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 $C_{11}\rtimes C_5$?

Thank you for your help!

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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 $-1$.

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  • $\begingroup$ Ahhhh thanks, this tiny mistake made me mad. $\endgroup$ – Phil Jan 4 at 12:00
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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

dieter.kaesbauer@googlemail.com

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