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Let $G$ be a finite abelian $p$-group with a unique subgroup $H$ of index $p$. It is a fact that $G$ is cyclic. This can be deduced from the classification theorem for finite abelian groups by writing $G$ as a product of cyclic groups and noting that $G$ does not have a unique subgroup of index $p$ unless there is only one such group. I am wondering if there is a nice proof of this result which does not use the classification theorem. Does anyone know of one?

Alexander Gruber
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user15464
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  • Take an element outside $H$, and consider the subgroup it generates... Note this does not need $G$ to be abelian. –  Feb 07 '13 at 07:09

3 Answers3

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If $G$ is finite then every proper subgroup is contained in a maximal subgroup, so every proper subgroup is contained in $H$. I claim therefore that $\langle x \rangle = G$ for any $x\in G\setminus H$ - why?

If $\langle x \rangle$ were proper in $G$, then we would have $\langle x \rangle \leqslant H$, a contradiction. So $\langle x \rangle=G$.

Note that this argument does not require that $G$ is abelian. In fact, we could even extend your hypothesis to "$G$ is a finite group with a unique maximal subgroup."

Alexander Gruber
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  • @user15464: Besides to the conclusion, we have $p\big||G/G'|$ as well. Maybe this fact be usefuul for the next. – Mikasa Feb 07 '13 at 07:17
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    It seems that this argument does not immediately extend to the non-abelian case. It is true that every maximal subgroup of a $p$-group has index $p$, but this requires proof. – Andrew Dudzik Sep 14 '14 at 01:49
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    Why every proper subgroup is contained in $H$?, $H$ is maximal, but is not obvious that $H$ is the unique maximal. For example, a group of index $p^2$ could be also a maximal subgroup. – Gaston Burrull Jun 12 '15 at 06:55
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Hint: It suffices to show that $G$ has precisely one subgroup of order $m$ for each $m$ dividing $|G|$. Show by induction that if we get more than one subgroup of order $p^k$ with $p^k\mid |G|$ then we can lift this to two distinct index $p$ subgroups. Thus, $G$ will have precisely one subgroup of order $p^k$ for each $p^k\mid|G|$ and thus $G$ will be cyclic.

Alex Youcis
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In fact the assumptions can be weakened and the conclusion will remain (almost) the same:

Let $G$ be a group of order $p^n$ for some prime $p$ and assume that $G$ contains a unique subgroup of order $p^k$ for some $1\leq k \leq n-1$.

Then either $G$ is cyclic or $p = 2$, $k = 1$ and $G$ is generalized quaternion (so $n\geq 3$).

For a proof of this, one can see Berkovich's Groups of Prime Power Order (it is Proposition 1.3).

Tobias Kildetoft
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