# It is sufficient to find the inverse of an element in a subgroup?

I have a simple proposition about groups, but it is not obvious for me. (This question is motivated by the last answer in Show identity of subgroup is same as identity of group by TheRealFakeNews)

Let $$(G, \circ)$$ and $$(H, +)$$ be groups with the identity elements $$1_G$$ and $$1_H$$, respectively. (The operation $$+$$ is defined to be the operation $$\circ$$ restricted on the elements in $$H$$. In other words, for all $$h_1, h_2 \in H$$ we have $$h_1 + h_2 = h_1 \circ h_2.$$)

Suppose that $$(H, +)$$ is a subgroup of $$(G, \circ)$$. Show that if $$h_1, h_2 \in H$$ and $$h_1 + h_2 = 1_H$$, then $$h_1 \circ h_2 = 1_G$$.

The proposition says that if $$h_2$$ is the inverse of $$h_1$$ in $$(H, +)$$, then also $$h_2$$ is the inverse of $$h_1$$ in $$(G, \circ)$$.

I know how to show that $$1_H = 1_G$$, from which the proposition is an easy consequence. But I want to prove the proposition, and use this to show that $$1_H = 1_G$$.

Since $$(H, +)$$ is a subgroup of $$(G, \circ)$$, we have $$h_1, h_2, 1_H \in G$$ and $$h_1 \circ h_2 = 1_H$$. Moreover, $$1_G \circ 1_H = 1_H \circ 1_G = 1_H$$ because $$1_G$$ is the identity in $$(G, \circ)$$. But I do now know to continue. Any suggestion please?

• NB: "proof" and "prove" are different words. Commented Sep 12, 2020 at 17:27

Let us denote the inverse of $$h$$ in $$H$$ by $$h^{(-)}$$, to possibly contrast with the inverse of $$h$$ in $$G$$, which is $$h^{-1}$$.
Let $$h\in H$$, and let $$h^{(-)}$$ be its inverse in $$H$$. Then you have that $$h(hh^{(-)}) = h$$ in $$H$$, but also in $$G$$. That means that $$hh^{(-)}$$ is an element of $$G$$ that satisfies the equation $$hx=h$$ in $$G$$. The only element that satisfies that equation in $$G$$ is $$1_G$$, so $$1_H = hh^{(-)} = 1_G$$ and $$h^{(-)} = h^{-1}$$.