H does not belong in this group. The alphabet is listing gradually down and H is out of place here.
bmsk
B
note b and g
Intro: g-g-a-a-a-a-a-a c-c-g-g-a-a-a Verse: g-g-g-g g-g-g-c g-g-g-g-g-g-g-g-g-a-b-a Refrain/Bridge: g-b-d-e-d-b-b-b-b-a g-b-d-e-d-b-a Chorus: b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-b-a-a b-b-a-g-a-b b-b-b-b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-a-b-a b-b-a-g-g-g-g-g-g-b-a-a b-b-a-g-a-b b-b-b-b-b-a-g-a-b b-b-b-b-b-a-g
Let H and I be subgroups of G. A group B is a subgroup of a group A if and only if every member of B is a member of A. Thus, every member of H is a member of G, and every member of I is a member of G. The intersection of two groups A and B is the set of things that are members of both A and B. Hence, if a group C is equivalent to the intersection of A and B, then everything that is a member of both A and B is a member of C, and everything that is a member of C is a member of both A and B. Hence, everything that is a member of the intersection of H and I is a member of H. We established above that every member of H is a member of G. Thus, everything that is a member of the intersection of H and I is a member of G. Recall that A group B is a subgroup of a group A if and only if every member of B is a member of A. Hence, if everything that is a member of the intersection of H and I is a member of G, then the intersection of H and I is a subgroup of G, and so the intersection of H and I is indeed a subgroup of G.
b a g a b b b a a a b b b b a g a b b b a a b a g b a g b a g g g g a a a b a g
easy-g a g g g g a g g g a g a a (2 times)-b b g g a a b g bb g g a a g -g a g g g g a g g g a g a a (2 times) hard-g a b b c b a g b b a g a a g a b b c b a g b b a g g a g d d b b a b c a b c d d a b c a b
The keyboard notes on the keyboard to the song, Call Me Maybe, are; B B B B B A B B B B B B A B B B B B B A A A A G G D B B B B B B A B B B B B B A B B B B B B A A A A G G D B D D B A G D D D B A G D D D B A G G G G G G G A B G B D G D B B D B G G B C B G G B A A G G B D G D B D B G G B C B G G B A A G G B D G D B B D B G G B C B G G B A A G G B D G G D B D B G G B C B G G B A A G.
BMSK does not belong because out of the four choices the other three all have spm in the group (meaning only one letter is different). BMSK only has MS in the group (meaning two letters are different because the P is missing)
G,b,a,g,g,b,a,g,c,b,a,a,a,g,b,a,g,g,b,a,g,g b,a,g,c,b,a,b,g
B b g a a g a a g e d b b g a a g a a g e d b b g a a g a a g e d d b g a g a g e a a g g g g d g g g g d a a b e d a a b e d g g g g d g g g d a a b e d a a b e b b b b b b a a a a a a a a g e d d g g g g d g g g g d a a b e d a a b e b b g a a g a g e d d b g a a g a a g e d d b b g a a g a a g e d d b b g a a g a a g e d d b g a g a g a g e a a g g g g d g g g g d a a b e d g g d g d g g g g d a a b e d a a b e b b b b b b a a a a a a a a g e d d g g g g d g g g g d a a b e d a a b e
For soprano recoders ,Jolly Old St. Nicholas goes like this:B-B-B-B-A-A-A-G-G-G-G-B-LOWE-LOWE-LOWE-LOWE-LOWD-LOWD-G-A-G-A-B-AB-B-B-B-A-A-A-G-G-G-G-B-LOWE-LOWE-LOWE-LOWE-LOWD-LOWD-G-A-G-A-B-G
E g e g e e g a b a g a / e g e g e e g a b a g a / e g e g e e g a b a g a e a b a / b a g / g g f f /f g a g g / b a g / g g f f / g a b g g g a b c2 / b b b g b g b b g b b b / g b g b b g / b b b g b g b g /c2 a g g g /c2 a g g g ................