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The question is: If $|G|=30$ and $|Z(G)|=5$, what is the structure of $G/Z(G)$?

I don't know what do we mean by 'structure' asked in the question. Please help.

Kushal Bhuyan
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Utkarsh
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    What is the behavior of the group elements under the operation? In essence, what is a common group it is isomorphic to? We know that all groups of order 6 are isomorphic to $S_3$ or $\mathbb Z / 6\mathbb Z$, and the G-Z theorem tells us what happens if $G/Z(G)$ is isomorphic to $\mathbb Z / 6 \mathbb Z$. – user217285 Dec 06 '15 at 09:29
  • Thanks @Nitin, for helping me out. – Utkarsh Dec 06 '15 at 10:09

1 Answers1

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$|G|=30$, $|Z(G)|=5$. $|G/Z(G)|=6$

We know that all groups of order 6 are isomorphic to $S_3$ or $\mathbb{Z} /6\mathbb{Z}$.

well known result: If $G/Z(G)$ is cyclic, then $G$ is abelian.

If $G/Z(G) \cong \mathbb{Z} /6\mathbb{Z} $ ,then $G$ is abelian which is contradiction to $|Z(G)|=5$ . Therefore $G/Z(G) \cong S_3$

user177523
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