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Let $H$ be a subgroup of $G$. Let $1_H$ and $1_G$ be the identities of $H$ and $G$, respectively. Show that $1_H=1_G$. My attempts is since we know that identity of a group is unique , and hence $1_H=1_G$. Is my proof true ?

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$1_H$ and $1_G$ are the identities of two different groups (albeit one is contained in the other), so the uniqueness of the identity of a group doesn't directly give that $1_G = 1_H$. – Brandon Carter Jan 31 '13 at 14:33
Hint: $1_G h=h $ for all $h \in H$. – PAD Jan 31 '13 at 14:40
up vote 9 down vote accepted

Working in $G$, we have $1_H1_H=1_H=1_H1_G$. The first equality follows from the fact that $1_H$ is the identity of $H$ and $H$ inherits its operation from $G$. The second follows from the fact that $1_G$ is the identity of $G$. Now premultiply by $1_H^{-1}$ to obtain the result.

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Start with $1_{H}^2 = 1_{H}$.

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The identity of $H$ is an element of $G$ satisfying $x^2 = x$, and the only such element can be $1_G$.

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Let $h_1, h_2 \in H$ where $h_2$ is the inverse of $h_1$. Note that this is allowed since $H$ itself is a group. Then $h_1h_2 = 1_H$, but $h_1, h_2$ are both elements in $G$ as well, so $h_1h_2 = 1_G$. Therefore, $1_H = 1_G$.

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