Let $G$ be a group and $\langle a \rangle$ be a unique subgroup of $G$ with order $n$. Then $a$ is in the center of $G$.

Question

I was working on my group theory exercises, and I encountered this problem.

"Let $G$ be a group and $a$ be a unique element of $G$ with order 2. Then $a$ is in the center of $G$."

My solution was considering a conjugation, which is an isomorphism. Since $a$ is the only element with the given property, it must be fixed by an isomorphism thus $gag^{-1} = a$ for all $g$, hence the result.

However, I want to generalize it as follows:

"Let $G$ be a group and $\langle a \rangle$ be a unique subgroup of $G$ with order $n$. Then $a$ is in the center of $G$."

I am stuck because I do not even know whether the statement is true or false. If this generalization fails to hold, for what $n$? Is there a condition for $n$, or $G$, to make this statement true?

Thanks in advance for all help, solution, or hint.

Answer 1

Consider the group $D_3 = \left<a,b:a^3= b^2=e,ab=ba^{-1}\right>$. The group has a unique subgroup of order $3$, given by $\left<a\right> = \{a,a^2,e\}$. However, $D_3$ has trivial center, so $a$ is not in the center.

Answer 2

You're close, but notice that conjugation doesn't quite work now!

Indeed, if $\langle a \rangle$ is the unique subgroup of order $n$, then you're right to notice that, acting by conjugation, $g^{-1} \langle a \rangle g = \langle a \rangle$ for every $g$, thus $\langle a \rangle$ must be normal in $G$.

However, that does not mean that $a$ is central in general. We know that $\langle a \rangle = \{ 1, a, a^2 ,\ldots, a^{n-1} \}$ is fixed setwise, but it can (and usually will) be permuted by the conjugation action. In the case $a$ has order $2$ then we know $\langle a \rangle = \{1 ,a \}$ must actually be fixed pointwise (do you see why?) so that $a$ is central, but for $n > 2$ there's enough wiggle room to allow for $a$ to not be fixed.

For instance, there's a unique subgroup of order $3$ in $S_3$, namely $\langle (1,2,3) \rangle = \{ \text{id}, (1,2,3), (1,3,2) \}$. Yet

$$(2,3)^{-1} (1,2,3) (2,3) = (1,3,2) \neq (1,2,3)$$

so that $(1,2,3)$ is not central.

Moreover, I don't know of a nice way to "salvage" this idea. In general, knowing that a subgroup is unique of a certain size tells you lots about it (for instance, it's always a characteristic subgroup), but it almost never buys you centrality.

---

I hope this helps ^_^

Answer 3

One of the generalizations runs as follows.

Proposition Let $N$ be a normal subgroup of a finite group $G$, such that $|N|=p$, the smallest prime dividing $|G|$. Then $N \subseteq Z(G)$.

Of course, if one would require $N$ to be unique in its order, then it would be normal.

To prove the proposition, note that a normal subgroup is always the disjoint union of some of the $G$-conjugacy classes. Since the conjugacy class of the identity is one of them, all the others must have a cardinality smaller than $p$. But that leaves a cardinality of $1$ for all others ($\#$class divides $|G|$). Hence $N$ is central.