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Let $Z(G)$ be the center of a non-commutative group $G$. Show that the factor group $G/Z(G)$ has at least 4 distinct subgroups.

Attempt:

So I have a guess at this however I feel that the answer is a bit too easy to be correct. So we have at least the two trivial subgroups, as we have to maintain our identity element under the inner automorphism, and the entire group. However since we in a sense 'kill' all the commuting elements by quotient by the centre we also have the two distinct subgroups of the left coset and the right coset.

So the four distinct subgroups are the identity, the entire group, and the left and right cosets. Is this correct or am I missing something?

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The left coset is not a subgroup, but only an element in the quotient group. As you notice that if the nontrivial quotient group has only two or three subgroups, then it is a cyclic group of order $p$ or $p^2$ ($p$ a prime), which make $G$ to be abelian, and $G=Z(G)$, a contradiction.

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  • $\begingroup$ A cyclic group of prime order has $2$ subgroups, not $3$. A cyclic group of order $p^2$ has $3$ subgroups, and cyclic groups of order $pq$ ($p,q$ distinct primes) or $p^3$ have $4$ subgroups. In fact non-cyclic groups must have at least $5$ subgroups. $\endgroup$ – Derek Holt Feb 9 '14 at 18:58
  • $\begingroup$ Yes, you are right. Corrected. Thanks. $\endgroup$ – Wei Zhou Feb 10 '14 at 0:06
  • $\begingroup$ What Theorem leads to the idea that a cyclic group of of order p squared has 3 subgroups? I can follow the logic, however I am finding difficulty linking the order of a quotient group to the number of subgroups inside it. $\endgroup$ – KaneZ Feb 10 '14 at 13:43
  • $\begingroup$ You can do it by yourself: if G is a cyclic group and k divide the order of G, then there exists exactly one subgroup of order k. Then ... $\endgroup$ – Wei Zhou Feb 10 '14 at 14:18
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Assuming the order of quotient group is finite, examine the cases in which $$\left|\frac{G}{Z(G)}\right|=1,2,3$$ and note that if $G/Z(G)$ is cyclic so $G$ is abelian.

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