# Let $G$ be a non-abelian group of order $10$. Prove that G has a trivial center.

I have done this as follows:

Let $$G$$ be a non-abelian group of order $$10$$.

If possible let a non-identity element say $$a \in G$$ is in $$Z(G)$$.

Now by lagrange's theorem , $$|a|= 2,5$$ or $$10$$.

If $$|a|=10$$

$$|a|$$ can't be $$10$$ (because then,$$G=$$, which will be a abelian group).

If $$|a|=2$$

Now, every non-identity element of $$G$$ will be of order $$2$$ or $$5$$. Now , their must be some element of $$G$$ of order $$5$$ (because if every non-identity element of $$G$$ be of order $$2$$, then $$G$$ will be abelian) say $$c \in G$$ , and $$|c| = 5$$. since, $$a \in Z(G)$$, so it will commute with every element of $$G$$, hence $$ac=ca$$ . Also $$|a|=2$$, $$|c|=5$$ . So $$|ac|=10$$. Which is again not possible since $$G$$ is non-abelian. Therefore, $$|a| \neq 2$$.

If $$|a|=5$$

Then $$a,a^2 ,a^3,a^4 \in G$$ , & all of these have order $$5$$ . So, in $$G$$ number of non-identity element of order $$5$$ will be $$4k$$ , where $$k \in \Bbb{N}$$. So if $$G$$ contains only non-identity element of order $$5$$, then order of $$G$$ will be of form $$4k+1$$ . But $$10$$ can't be of form $$4k+1$$. So there must be a non-identity element of order $$2$$ say $$d$$. Again by similar argument as above $$|ad| = 10$$ , which is again an impossibility. So$$|a| \neq 5$$.

Therefore our assumption that $$a$$ is a non-identity element that is in $$Z(G)$$, is wrong. So$$Z(G)$$ has only identity element. So it is a trivial subgroup of $$G$$.

If any mistakes please correct. And if you could provide a better solution than, please do.

• If $G/Z(G)$ is cyclic then $G$ is abelian May 9, 2019 at 18:22
• May 9, 2019 at 18:35

## 2 Answers

Fact. If $$G/Z(G)$$ is cyclic then $$G$$ is abelian. For a proof, see the answers to this old question.

As in the question $$G$$ is non-abelian, the above fact means that $$G/Z(G)$$ is non-cyclic.

Suppose that $$Z(G)$$ is non-trivial. Then $$G/Z(G)$$ has order either $$2$$ or $$5$$ (why?). Hence, $$G/Z(G)$$ is cyclic (why?), a contradiction. Hence, $$Z(G)$$ is trivial as required.

Any non-abelian group of order $$2p$$ is isomorphic to the dihedral group $$D_p$$. For $$p=5$$ we obtain $$D_5$$, which has trivial center.

References:

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .

Center of dihedral group