# Generators of $\mathbb{Z}_{5}\times\mathbb{Z}_{11}$

My task was to find all groups of order 55 and then describe each of them with generators. I found that there is only 1 abelian group, $$\mathbb{Z}_{5}\times\mathbb{Z}_{11}$$. I don't know how to find generators of this group. Any hints?

P.S. Also I found that there is only 1 non-abelian group with generators: $$G = C _ { 11 } \times _ { \phi } C _ { 5 } = \langle x , y | x ^ { 11 } = 1 , y ^ { 3 } = 1 , y x = x ^ { 4 } y \rangle$$

## 3 Answers

Hint: this group is isomorphic to $$\mathbb{Z_{55}}$$.

• So: $x ^ { 11 } = 1 , y ^ { 5 } = 1 , y x y ^ { - 1 } = x ^ { r } , \text { for some } r , 1 \leq r < 11$? – Chyma Feb 17 at 18:03
• You wrote your task was to find all the groups of order $55$ up to isomorphism. So instead of $\mathbb{Z_5}\times\mathbb{Z_{11}}$ you can take the group $\mathbb{Z_{55}}$ as they are isomorphic according to the Chinese remainder theorem. And it is obvious that $\mathbb{Z_{55}}=\langle 1\rangle$. If for some reason you still want to work with $\mathbb{Z_5}\times\mathbb{Z_{11}}$ you can prove the element $(1,1)$ has order $55$, hence generates the whole group. – Mark Feb 17 at 18:06

You can work with $$\Bbb Z_{55}$$ and there will be $$\varphi (55)=40$$ generators, corresponding to integers coprime with $$55$$.

Since $$5$$ and $$11$$ are relatively prime, the element $$(1_5, 1_{11})$$ will generate $$\mathbb{Z}_5\times \mathbb{Z}_{11}$$