Question

1. Let G - Z. Let H - {0,3,5,9) be a subgroup of (you do not need to prove this is a subgroup of G). Prove that G/l is a valid quotient group. Explain what the elements of G/H are and what the group operation is. 2. Let G be a group and H a normal subgroup in G. I E H for all IEG, then prove that G/H is abelian

Answer 1

(+) set G-212. Let H = { 0, 3, 5, §} be a subgroup of G. Prove that G/H is a valid quotlent group. Explain what the elements of G/H are and what the group operation is? solution since G= 2 is an Abelian group. therefore every sub group of G is normal in G. H = 45, 3, 5, §} is also normal in G. in G is valid quotient group. (If G is a group. His a normal subgroup of G. Then G is valid Next, operation on G = 242 Sfatu: ae G+2,2} well defined G=242 = { 0 , I , 2, 3, 4, 5, 6, 7, 8, 9, 10, II } : 0 + H = { 0+, 0+3, 7+6, +9)= {0,3,6,9}=H. 1+H = { Ito, It, It, Itğ } = {1, a, 7, 10} 2+H = { 2 to, ž+, Itt 2+ ] } = {2, 5, 5, II } 3+H - {3 to, 3+3, 3+5, 3+3 } = {3, 6, 9, 12:0} =H= 5+1H 8th similarly, A+H = itH = 7+H = 10+H. 5+H - 2+ - +H = 11 + H. H Hence, a = {H, ItH, Z+H.}. Operation on G. (a+h)* 16+H) = (a+b)+H. (*)

According to operation on GI LH : 0 + H. (Ô+ H) x (0 + H) - (0 +) + 1 = 0 + H (5+H) (I+H) - (5+1)+H (1 +H) (1+H) = (1 + i) +H (I+H) & (+H) = 1 + 2) +H (2 + H) (2+H) = (2+2) + H - ItH. - Z+H - 3+H = 5+H. 4 +H - ItH . and so on. (This is just for your better understanding). it (2). Let G be a group and H is a normal subgroup in G. - G is well defined. Also VX EG x²EH (given). To Prove G is Abelian. Phu : { $H : xe GÌ operation on G case I: If XEH, then Xo = H ( Identity of G). Case I set XeG but x&H - operation on a Then (XH)² = (x+1)+(x-H) = (x-x)•H = x² •H = H (since for all & EG, x²EH) - Identity of G

I order of x-H) is either 1 or 2. of olxH) = 1 :0(a) = 1 iff a= identity/ VOEG. XH = H (Identity of G EXEH ( XHEHEH) VXEG. (but from (* XEGbut X&H) Hence it is a contradiction. - 01xH) can not be one. =) Olx H) = 2 . since (XH) is arbitrary elt. of G. therefore order of every non-identity element is 2. each element in G is self - inverse. – (**) ie. (2 H)2 = x.H. Now, (x-4) + (8-4)= (2+1 (9-4)) (8.4) * (**)) = H ( order of each elt, in => (X-H) * (9-H) = (x H) + (y:-))-2 (9.- *()-2 (lab) = 6+a+) = (y.H) & (x-H) (by & ) Vĩ H, 18 +H) + . is Abelian.
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