0
$\begingroup$

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: $G = C _ { 11 } \times _ { \phi } C _ { 5 } = \langle x , y | x ^ { 11 } = 1 , y ^ { 3 } = 1 , y x = x ^ { 4 } y \rangle$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So: $x ^ { 11 } = 1 , y ^ { 5 } = 1 , y x y ^ { - 1 } = x ^ { r } , \text { for some } r , 1 \leq r < 11$? $\endgroup$ – Chyma Feb 17 at 18:03
  • $\begingroup$ 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. $\endgroup$ – Mark Feb 17 at 18:06
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.