how to show D is cyclic knowing it has a unique subgroup?

Question

If D has $25$ elements with a unique subgroup of order $5$ , show that $D$ is cyclic ?? I know that every cyclic group has a unique subgroup . How does the converse work here ???

Answer 1

Let $H$ be the unique subgroup of order $5$. Then $H$ has $5$ elements. Take $g \in D \setminus H$. Then the order of $g$ divides $25$ and cannot be $1$ or $5$. Therefore, it is $25$ and $D$ is cyclic generated by $g$.

More generally, the same argument proves this:

If $G$ is a finite group of order $p^n$ and has at most one subgroup of each possible order, then $G$ is cyclic.

Indeed, the set of elements of order less than $p^n$ has size at most $$ 1 + p + p^2 + \cdots + p^{n-1} = \frac{p^n-1}{p-1} < \frac{p^n}{p-1} < p^n $$

Answer 2

Since every group of order $25$ is always abelian.So, $G$ can be of the form ${\Bbb Z}_5 \times {\Bbb Z}_5$ or of the form ${\Bbb Z}_{25}$. Now since the subgroup has unique subgroup of order $5$,we must show that $G$ will be cyclic i.e. ${\Bbb Z}_{25}$. Clearly we can see that since ${\Bbb Z}_5 \times {\Bbb Z}_5$ has not unique subgroup of order $5$. So We are done.

But If you want the proof in a fundamental way,then here is the proof:

Let $x\in G$, $x\neq e$.Now the order of $x$ is either $5$ or $25$; if it is $25$, we are done, because then $\langle x\rangle = G$ and $G$ is cyclic. So assume that $x$ has order $5$. Then $H=\langle x\rangle$ is the unique subgroup of $G$ of order $5$, by assumption. Now, $G-H$ is not empty ; let $y\in G-H$. Again, the order of $y$ must be either $1$, $5$, or $25$: it cannot be $1$,because the identity is in $H$, whereas $y\in G-H$.

The order cannot be $5$ either: if the order were $5$, then $K=\langle y\rangle$ would be a subgroup of $G$ of order $5$ but by our assumption $G$ has a unique subgroup of order $5$, so that would mean that $K=H$ since both have order $5$; but then $y\in \langle y\rangle = K = H$, again contradicting our choice of $y$ as an element of $G-H$.

But that only leaves one possibility: that the order of $y$ is $25$. But if the order of $y$ is $25$, then $\langle y\rangle = G$, so $G$ is cyclic, as claimed.