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I'm not sure I've convinced myself here.

Take the L be the set of all linear transformations from V -> V, where V is a finite dimensional vector space. We know that G contained within L of all invertible linear transformations is a group. Say you have S, T within L and ST = I [ST is functional composition S(T(v))], where I is the identity mapping. Then it must be that ST is invertible (since we know that 1-1 <=> onto <=> invertible in L) since ST is a 1-1 and onto mapping. Which means ST in G and the inverse of (ST)^-1 = T^-1S^-1.....but this also implies that T and S are invertible and thus in G??????

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# If ab in G, then a and b in G?

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