Any help on how to prove the following:
If a group $G$ is of order $42$ with center of size $6$, then $G$ is abelian.
Asked
Active
Viewed 105 times
-2
Chuks
- 1,261
- 7
- 18
-
3"need some help to prove that...." - so you tried it yourself and got stuck somewhere, right? Can you please elaborate and share your attempts? – barak manos Jan 10 '16 at 10:45
3 Answers
5
Notice that $G/Z(G)$ is of order $7$ and thus cylic, so the statement follows from the fact that $G$ is abelian if $G/Z(G)$ is cylic.
I think it is interesting to notice that formally speaking, this shows that no such group $G$ exists, because $G$ being abelian implies $Z(G) = G$, so $Z(G)$ is of order $42$ and not $6$. So the statement, that every such group $G$ is abelian, is true because not such group exists.
Jendrik Stelzner
- 17,081
-
you've opened my mind to fact that the order of $G$ and $Z(G)$ are different Although $|G:Z(G)|=7$. Is that means that the question is wrong?? – Lauren Jan 10 '16 at 13:57
-
The statement of the question is right: Every group $G$ with $|G|=42$ and $|Z(G)| = 6$ is abelian. It is just that there exists no such group. So every statement about such a group $G$ is true. (I feel like this is more of a technicality then the indented answer to the question.) – Jendrik Stelzner Jan 10 '16 at 14:17
-
so if such a group does not exist can we add any statement we want(for exp it is polycyclic)??? – Lauren Jan 10 '16 at 15:08
-
Yes, because for every group $G$ the implication "($|G| = 42$ and $|Z(G)| = 6$) $\implies$ statement" holds. – Jendrik Stelzner Jan 10 '16 at 15:13
-
1
First $Z(G)\trianglelefteq G$. Now consider $G/Z(G)$. We have that $|G/Z(G)|=\frac{|G|}{|Z(G)|}=\frac{42}{6}=7$.
Up to isomorphism, $\mathbb{Z}_7$ is the only group of order $7$. So $G/Z(G)\cong \mathbb{Z}_7$.
There is a lemma that says that if $G/Z(G)$ is cyclic, then $G$ is abelian. The result follows from this lemma.
Chuks
- 1,261
- 7
- 18