There is a simple argument that shows that any two subgroups of a cyclic group that are isomorphic must be identical. This is because they can each be represented in terms of the generator of the cyclic group.

This made me wonder, what are all finite groups with the property that any two isomorphic subgroups are identical? My conjecture is that they must all be cyclic. Is this correct?

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    $\begingroup$ It certainly means that every subgroup is normal, due to the conjugation isomorphisms. $\endgroup$ Commented Mar 13, 2018 at 17:36
  • $\begingroup$ Yes, right now I am stuck trying to prove all Sylow p-subgroups are cyclic. I think that is enough to prove my conjecture. $\endgroup$
    – abnry
    Commented Mar 13, 2018 at 17:37
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    $\begingroup$ A non-cyclic abelian group doesn't have this property: if we write the group as a product of prime power cyclic groups, there must be some prime $p$ which appears twice, so we have a subgroup isomorphic to $\Bbb Z_p\times \Bbb Z_p$, and therefore many different order-$p$ subgroups (which must all be isomorphic). So we're left with the non-abelian possibility. $\endgroup$
    – Arthur
    Commented Mar 13, 2018 at 17:40

2 Answers 2


Let $G$ be a finite group with this property. Let $a\in G$ and let $a$ have order $k$. Then every element of order $k$ in $G$ generates $\left<a\right>$. So there are exactly $\varphi(k)$ elements of order $k$ in $G$. So for each $k$, $G$ has either $0$ or $\varphi(k)$ elements of order $k$.

Let $|G|=n$. Then for each $k$ for which there are elements of order $k$ in $G$, then $k\mid n$. But $n=\sum_{k\mid n}\varphi(k)$. If for some $k\mid n$ there are no elements of order $k$ in $G$, then the total number of elements in $G$ is $<n$, which is impossible.

In particular, $G$ has an element of order $n$. Thus $G$ is cyclic.


From the comments, here is short answer.

As pointed out, any exception to the conjecture must be a group that is non-abelian /and/ has every subgroup normal.

This means the group has the property of being a Dedekind group. In particular, there is a result that the quaternion group $Q_8$ is a subgroup of all such groups. The quaternion group as 6 elements of order four, which cannot belong to a single cyclic group of order four.

We conclude such an exception is not possible.

This depends heavily of the properties of a Dedekind group, which I have no insight to. A better answer would be much appreciated.

  • $\begingroup$ The quaternion group has a unique element of order $2$. $\endgroup$
    – verret
    Commented Mar 13, 2018 at 18:21
  • $\begingroup$ The presentation of the quaternion group has $i^2=j^2=k^2=e$, with $i, j, k$ unique. en.wikipedia.org/wiki/Quaternion_group $\endgroup$
    – abnry
    Commented Mar 13, 2018 at 18:50
  • $\begingroup$ No, in the usual terminology, it has $i^2=j^2=k^2=-1$, with $(-1)^2=1$. On the page you are citing, they are calling $-1=\overline{e}$. (Note the distinction between $e$ and $\overline{e}$!) $\endgroup$
    – verret
    Commented Mar 13, 2018 at 21:41
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    $\begingroup$ Anyway, the argument in your post can still be fixed, because $Q_8$ has three distinct cyclic subgroups of order $4$. $\endgroup$
    – verret
    Commented Mar 13, 2018 at 21:42
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    $\begingroup$ The fix is still not clear. Note that each subgroup of order $4$ has two elements of order $4$, so in principle, one could have two subgroups of order $4$ and four elements of order $4$. Anyway, the quaternion group has exactly six elements of order $4$, and three cyclic subgroups of order $4$. $\endgroup$
    – verret
    Commented Mar 13, 2018 at 21:54

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