If g and h are of same order in a finite group G, then there exists a group H, an extension of G such that g and h are conjugates in H .

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Actually, we have seen restriction of every structural space . And, there are extensions also, like extension of a metric space into another .

I thought about extending a group, but I can't figure out how to implement closure operations and inverses .

And the question : Is the above statement true ??? If true, then how to relate the finiteness of G with it ???

[Consequences of Britton's Lemma
Edit Most basic properties of HNN-extensions follow from Britton's Lemma. These consequences include the following facts:

The natural homomorphism from G to {\displaystyle G_{\alpha }} {\displaystyle G_{\alpha }} is injective, so that we can think of {\displaystyle G_{\alpha }} {\displaystyle G_{\alpha }} as containing G as a subgroup. Every element of finite order in {\displaystyle G_{\alpha }} {\displaystyle G_{\alpha }} is conjugate to an element of G. Every finite subgroup of {\displaystyle G_{\alpha }} {\displaystyle G_{\alpha }} is conjugate to a finite subgroup of G. If {\displaystyle H\neq G} {\displaystyle H\neq G} and {\displaystyle K\neq G} {\displaystyle K\neq G} then {\displaystyle G_{\alpha }} {\displaystyle G_{\alpha }} contains a subgroup isomorphic to a free group of rank two.]- source HNN extension - Wikipedia

But, the proof of Britton's lemma is not provided, then how to approach ???

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There's the HNN extension. Let $a$ and $b$ be the elements in question and let $H$ be given by the the following presentation. Define $G$ by generators and relations, add a new generator $t$ and a new relation $at=tb$. One problem is that the HNN extension in general has infinite index, so if $G$ is finite then in general $H$ won't be.