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So, a single element normal subgroup $n$ of group $G$ would be defined as $ \forall g \in G: gng^{-1} = n$. All elements of an abelian group would qualify, and so would more generally the elements of the center of a group, since $gng^{-1} = gg^{-1}n = n$ for $n\in \text{Z}(G)$.

Thus, to prove that these are the only elements for which this is true, we consider $gng^{-1} = n$. Right multiplying by $g$ yields $gng^{-1}g = ng$, which simplifies by associativity to $gn = ng$, implying that $g$ and $n$ commute. Thus, $n\in \text{Z}(G)$.

Is this line of reasoning correct?

Edit: My terminology was confusing in calling things "single element normal subgroup". They are not subgroups. I think a better question would have been to prove that the only normal subgroups $N$ where $\forall n \in N, \forall g \in G: gng^{-1} = n$ are subgroups of the center of $G$.

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    $\begingroup$ I wonder if what you are thinking about are conjugacy classes. Central elements are only conjugate to themselves and the union of all these singleton conjugacy classes makes up the centre of the group. $\endgroup$ Jun 23 '19 at 17:33
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The phrase "single element normal subgroup" is confusing. The only subgroup with a single element is the subgroup consisting only of the identity.

Your argument correctly proves that the center of the group is the set of elements fixed by every inner automorphism.

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The only issue, as mentioned by others, is that the only single element subgroup is the trivial group. But it's a common exercise along these lines to show normal subgroups of order $2$ are in the center for the reasons you say.

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The only single element (normal) subgroup of a group is $\{e\} $, where $e $ is the identity element. No other single element subgroup is possible.

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If $g$ is the unique element of order $n$, then $n=1$ or $n=2$ and $g$ must be central. Perhaps this is what you mean, Leo?

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As others have pointed out, every subgroup must contain the identity, so there is only one single-element subgroup, namely $\{1_G\}$.

However, I think there is a way to rescue this claim (and its proof) in a non-trivial way. Define a normal subset to be a subset $S\subseteq G$ with the property that $gS = Sg$ for all $g\in G$. Note, this is not standard terminology; however, it does work well with the standard phrase "normal subgroup", in that a normal subgroup would just be a subgroup that is a normal subset in this sense.

What you have actually proved is that the only single-element normal subsets are precisely the elements that make up the center of a group.

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