I have the following (from Aluffi II.7.7):
Theorem: Let $G$ be a group and $H \leq G$ be generated by all elements of a fixed order $N$. Then $H$ is a normal subgroup of $G$.
I came up with a fairly simple proof using the concrete description for "the subgroup generated by a subset": the subgroup consists of exactly the products of elements of the subset.
Proof: Using the concrete description for $H$, any element $h \in H$ can be written as: $$ h = \prod_{1 \leq i < k} a_i,$$ where each $a_i$ has order $N$. But for any $g \in G$, we know each $g a_i g^{-1}$ has order $N$ as well, so: $$ \prod_{1 \leq i < k} ga_ig^{-1} \in H,$$ but this product is just $ghg^{-1}$. Therefore, $H$ is normal in $G$.
However, I was wondering if there's a proof that uses the following definition of "the subgroup generated by a subset":
Definition: For any $A \subseteq G$ with the natural map $\iota: A \to G$, the universal property for free groups implies that there exists a unique $\varphi: F(A) \to G$ such that the following diagram commutes:
The subgroup generated by $A$ is then the image of $\varphi$.
One of my attempts is as follows:
Let $G$ be a group, and let $A$ be the elements of order $N$ in $G$. For any $g \in G$, let $\gamma_g$ denote conjugation by $g$; consider the following diagram:
The universal property for free groups guarantees that $\varphi_g$ exists and is unique. However, we also have $\varphi: F(A) \to G$ from the definition of the subgroup generated by $A$:
...
I'm convinced that $??? = \gamma_g$ makes the diagram commute, in which case for any $h \in \text{img }\varphi$, then $h = \varphi(a)$ for some $a \in F(A)$ and $$ghg^{-1} = \gamma_g(\varphi(a)) = \varphi(\varphi_g(a)).$$ Thus, $ghg^{-1} \in \text{img } \varphi$. I'm stuck on how to prove that $??? = \gamma_g$ with just universal properties. I tried invoking the universal property with $\varphi \, j \, \gamma_g : A \to G$ but couldn't seem to get anywhere. Does anyone have suggestions on how to proceed from here?
As David pointed out in the comments, "proving that ??? $= \gamma_g$" is the wrong way to think about it. In fact, there can be other endomorphisms of $G$ that make your diagram commute; for example, in the silly case $N = 1$, any endomorphism of $G$ makes the diagram commute.
Instead, I would go back to the definition of the morphism $\gamma_g: A \to A$. By definition, this is the restriction of $\gamma_g: G \to G$ to the subset $A$. It follows that the outer rectangle of your diagram commutes, with ??? = $\gamma_g$. The problem then is to prove that the upper square commutes as well. That is, we must show that $\gamma_g \varphi = \varphi \varphi_g: F(A) \to G$. By (the uniqueness part of) the universal property of free groups, it suffices to prove that $\gamma_g \varphi j = \varphi \varphi_g j: A \to G$. This follows from the commutativity of the bottom square and the outer rectangle.