# Non abelian finite group of order $55$

Give an example of a non abelian group of order $$55$$.

To find non abelian group the simplest way is to find one non abelian group whose order divides the order of given group and then we take the group which is the external direct product of the non abelian group and some other abelian group.

For example to find a non abelian group of order $$36$$ we take the permutation group $$S_3$$ and take the group $$S_3\otimes \Bbb Z_6$$. But using this way we can not have a group of order $$55$$ since any group of order $$5$$ or $$11$$ will be abelian.

So how do we proceed?

• Semi-direct products? – Angina Seng Aug 18 '19 at 15:47

Notince that $$\operatorname{aut}\Bbb Z_{11}\cong \Bbb Z_{11}^*\cong \Bbb Z_{10}$$. Therefore you can consider an element $$g\in\operatorname{aut}\Bbb Z_{11}$$ of order $$5$$ and the homomorphism $$\phi:\Bbb Z_5\to \operatorname{aut}\Bbb Z_{11}$$, $$\phi(m)=g^m$$. Then you have a non-abelian semidirect product $$\Bbb Z_{11}\rtimes_\phi \Bbb Z_5$$ with operation $$(a,b)*(c,d)=(a+\phi(b)(c),b+d)$$.
• Another way to describe the resulting group in this case is as a subgroup of all affine maps from $\mathbb{Z}_{11}$ to itself. – Tobias Kildetoft Aug 18 '19 at 15:58
• (In fact, the subgroup comprising the affine maps $t \mapsto a t + b$ for $a$ a nonzero square in $\Bbb F_{11}$, i.e., $a \in \{1, 3, 4, 5, 9\}$.) – Travis Willse Aug 18 '19 at 19:33
Consider the field $$G=\Bbb Z_{11}$$. Now its multiplicative group $$\Bbb Z_{11}^*$$ is a group of order $$10$$. Now $$5$$ divides $$10$$, so By Cauchy's theorem, there exist a subgroup $$H$$ of order $$5$$ in $$\Bbb Z_{11}^*$$. Now consider $$S=\left\{ \begin{pmatrix} 1& x \\ 0& y \end{pmatrix} : x \in \Bbb Z_{11}, y \in H\right\}$$ Then $$S$$ is a non abelian group under matrix multiplication of order $$55$$
By Sylow's Third Theorem the number $$n_{11}$$ of Sylow $$11$$-subgroups of a group $$G$$ of order $$55$$ divides $$5$$ and is congruent to $$1$$ modulo $$11$$, so any group of order $$55$$ has one subgroup $$N$$ of order $$11$$, which we just call $$\Bbb Z_{11}$$. Pick any subgroup $$H$$ of order $$5$$ (which exists by Cauchy's Theorem); we likewise call it $$\Bbb Z_5$$. Since $$5$$ and $$11$$ are coprime, $$\Bbb Z_5 \cap \Bbb Z_{11} = \{1_G\}$$ and thus $$\Bbb Z_{55} = \Bbb Z_{11} \Bbb Z_5$$, so our group is isomorphic to a semidirect product: $$G \cong\Bbb Z_{11} \rtimes \Bbb Z_5 .$$ In other words, there is some homomorphism $$\phi : \Bbb Z_5 \to \operatorname{Aut}(\Bbb Z_{11}) \cong \Bbb Z_{10}$$ such that $$G$$ is isomorphic to the set $$\Bbb Z_{11} \times \Bbb Z_5$$ equipped with the group operation $$(n, h) \ast_\phi (n', h') \mapsto (n + \phi(h)(n), h + h') .$$ Since $$\Bbb Z_5$$ is cyclic, $$\phi$$ is determined by $$\phi(1)$$ and $$\phi(1) = n$$ defines a group operation iff $$n$$ has order dividing $$5$$. The choice $$\phi(1) = 0$$ determines the trivial homomorphism $$\phi$$ and thus the direct product $$\Bbb Z_{11} \times \Bbb Z_5$$. All of the other choices are conjugate and so define isomorphic semidirect products, so we need consider only one, say, the one defined by $$\phi(1) = 2$$, which corresponds to the homomorphism $$\phi : n \mapsto n^2$$, and this defines a nonabelian group operation $$\ast_\phi$$, $$(n, h) \ast\phi (n', h') = (n + (n')^{2h}, h + h').$$ (In fact we have shown that up to isomorphism the group defined by this operation and the cyclic group of order $$55$$ are the only groups of that order.)
We can interpret this group concretely: Consider the group $$\operatorname{Aff}(\Bbb F_{11})$$ (which has order $$110 = 55 \cdot 2$$) of invertible affine transformations $$t \mapsto a t + b$$ of $$\Bbb F_{11}$$. The composition rule is $$(t \mapsto a t + b) \circ (t \mapsto a' t + b') = (t \mapsto (a a') t + (a b' + b)) ,$$ and so we can identify this group as the semidirect product $$\Bbb Z_{11} \rtimes \operatorname{Aut}(\Bbb Z_{11}) \cong \Bbb Z_{11} \rtimes \Bbb Z_{10} .$$ Thus, we can identify our nonabelian group $$G$$ of order $$55$$ with the subgroup of $$\operatorname{Aff}(\Bbb F_{11})$$ whose elements have $$\operatorname{Aut}(\Bbb Z_{11})$$-component of order dividing $$5$$, that is, whose elements have $$\Bbb Z_{10}$$-component contained in the unique copy $$\Bbb Z_5 \cong \{0, 2, 4, 6, 8\} < \Bbb Z_{10}$$. So, these are just the squares in $$\operatorname{Aut}(\Bbb Z_{11})$$, and so we thus recover an observation of Tobias Kildetoft from a comment about another answer: We can identify $$G$$ up to isomorphism as the group affine transformations $$t \mapsto a t + b$$ of $$\Bbb F_{11}$$ with $$a$$ a square, that is, $$G \cong \{t \mapsto a t + b : a \in (\Bbb F_{11}^\times)^2, b \in \Bbb F_{11}\} = \{t \mapsto a t + b : a \in \{1, 3, 4, 5, 9\}, b \in \Bbb F_{11}\} .$$ Applying the usual embedding $$\operatorname{Aff}(\Bbb F) \hookrightarrow \operatorname{GL}(2, \Bbb F)$$ then recovers the explicit matrix group realization in Chinnapparaj R.'s answer, $$G \cong \left\{ \pmatrix{1 & b \\ & a} : a \in (\Bbb F_{11}^\times)^2, b \in \Bbb F_{11} \right\} \subset \operatorname{GL}(2, \Bbb F_{11}).$$