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=<a> $, 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.

  • 5
    $\begingroup$ If $G/Z(G)$ is cyclic then $G$ is abelian $\endgroup$
    – Lozenges
    May 9, 2019 at 18:22
  • 2
    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    May 9, 2019 at 18:35

2 Answers 2


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.


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

Center of dihedral group


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