I've tried to prove that the number of left and right cosets of a subgroup should be the same.
Statement: if [imath]H[/imath] is subgroup of [imath]G[/imath], then number of left cosets of [imath]H[/imath] equals number of right cosets of [imath]H[/imath].
Proof: suppose the opposite - number of left cosets (later just [imath]C_l[/imath]) equals [imath]n[/imath] but number of right cosets (later [imath]C_r[/imath]) equals [imath]n+i[/imath] ([imath]n, i \in \mathbb N[/imath]). This means that there is a subset of [imath]G[/imath], [imath]I=[g_1, g_2,..., g_i][/imath], such that each left coset of the form [imath]g_jH=k_jH (\forall g_j \in I)[/imath], (where [imath]k_j[/imath] denotes some other element of [imath]G[/imath]) though right cosets of the form [imath]Hg_j[/imath] are all unique (by "unique" I mean that there is no such [imath]k_j \in G[/imath] satisfying [imath]Hg_j=Hk_j[/imath]). Now lets take all those equations of the form [imath]g_jH=k_jH[/imath] and multiply them by [imath]g_j^{-1}[/imath] on the left. We now have that [imath]H=g_j^{-1}k_jH[/imath], which can mean either that 1)[imath]g_j^{-1}k_j=e_G[/imath] or 2) [imath]g_j^{-1}k_j \in H[/imath]. In the first case, we have [imath]g_j^{-1}k_j=e_G[/imath], which means that [imath]k_j=g_j[/imath], so all left cosets of the form [imath]g_jH[/imath] are now unique (because the only [imath]k_j \in G[/imath] satisfying [imath]g_jH=k_jH[/imath] is [imath]g_j[/imath] itself). Because there are exactly [imath]i[/imath] elements in [imath]I[/imath], and all of their left cosets are unique, [imath]C_l=n+i=C_r[/imath]. In the second case, having all [imath]g_j^{-1}k_j[/imath] in [imath]H[/imath] also gives us that all of the elements in [imath]I[/imath] are now in [imath]H[/imath] (because [imath]H[/imath] is a subgroup). But that implies that all right cosets of the form [imath]Hg_j[/imath] are the same and cannot be unique. There were [imath]i[/imath] of these cosets in total, so [imath]C_r[/imath] must decrease exactly by [imath]i[/imath] ([imath]C_r-i=(n+i)-i=n=C_l[/imath]).
Even if my proof is correct, please tell if there's some way to make it shorter.
Statement: if [imath]H[/imath] is subgroup of [imath]G[/imath], then number of left cosets of [imath]H[/imath] equals number of right cosets of [imath]H[/imath].
Proof: suppose the opposite - number of left cosets (later just [imath]C_l[/imath]) equals [imath]n[/imath] but number of right cosets (later [imath]C_r[/imath]) equals [imath]n+i[/imath] ([imath]n, i \in \mathbb N[/imath]). This means that there is a subset of [imath]G[/imath], [imath]I=[g_1, g_2,..., g_i][/imath], such that each left coset of the form [imath]g_jH=k_jH (\forall g_j \in I)[/imath], (where [imath]k_j[/imath] denotes some other element of [imath]G[/imath]) though right cosets of the form [imath]Hg_j[/imath] are all unique (by "unique" I mean that there is no such [imath]k_j \in G[/imath] satisfying [imath]Hg_j=Hk_j[/imath]). Now lets take all those equations of the form [imath]g_jH=k_jH[/imath] and multiply them by [imath]g_j^{-1}[/imath] on the left. We now have that [imath]H=g_j^{-1}k_jH[/imath], which can mean either that 1)[imath]g_j^{-1}k_j=e_G[/imath] or 2) [imath]g_j^{-1}k_j \in H[/imath]. In the first case, we have [imath]g_j^{-1}k_j=e_G[/imath], which means that [imath]k_j=g_j[/imath], so all left cosets of the form [imath]g_jH[/imath] are now unique (because the only [imath]k_j \in G[/imath] satisfying [imath]g_jH=k_jH[/imath] is [imath]g_j[/imath] itself). Because there are exactly [imath]i[/imath] elements in [imath]I[/imath], and all of their left cosets are unique, [imath]C_l=n+i=C_r[/imath]. In the second case, having all [imath]g_j^{-1}k_j[/imath] in [imath]H[/imath] also gives us that all of the elements in [imath]I[/imath] are now in [imath]H[/imath] (because [imath]H[/imath] is a subgroup). But that implies that all right cosets of the form [imath]Hg_j[/imath] are the same and cannot be unique. There were [imath]i[/imath] of these cosets in total, so [imath]C_r[/imath] must decrease exactly by [imath]i[/imath] ([imath]C_r-i=(n+i)-i=n=C_l[/imath]).
Even if my proof is correct, please tell if there's some way to make it shorter.
