Mathematics Asked by Kapur on November 17, 2020

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 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?

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 $hin H$, 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 $h^{(-)} = h^{-1}$.

Correct answer by Arturo Magidin on November 17, 2020

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