Question

Let o and be the following symmetries of the regular pentagon: 240° rotation reflection 5 6 (a) (6 points) Express o and T as permutations of the vertices (i.e., in two-row notation). (b) (9 points) Compute the following permutations. Express answers in two-row notation. i. OOT ii. τοσ iii. TOOOT (c) (5 points) Determine the symmetry represented by TOOOT. Give the axis of symmetry if a reflection, or the degrees of clock-wise rotation if a rotation.

Answer 1

Here is the following symmetry 6 2 2 4 240* rotation 3 5 5 4 6 And e is the reflection 2 6 2 3 reflection 5 て 3 4 4 5

After 240 notation, le inr rymmary 1 +5 276 3 1 1 4 -12 5 + 3 6 (1) - 5. ()=6, (3) = 1, + (4) = 2, (5) (6) = 4. = 4 le -3, → 11 1 4 3 2 6 4 2 5 6 3 4 5 For てい le after the reflection about the which is through midpoint of verteces. and 2 QXS line passing 5 and Thus. 1 + 2 : 0) = 2 2 + 1 i {(2)= 1 36 ?(3)=6 4 >5 (4)=5 5 54; ?(5)= 4 6- 3 i (6)=3

Thus て -C 2 - a 3 6 4 5 5 6 4 3 ) 2 - b 0 Want to compute. 20s Ae we know rC)= 5 CO= 2 r(2) = 6 +(2)=1 (3) T(3) = 6 r(4) = 2, (4) - 5 (5) = 3 E(5)= 4 + (6) = 4, (6)= 3 → rote 1) = (t())>(2) = 6 To(2) = ( (2)) ( )= 5 → rot(2)=5 rof(3) = ( (3) (-4, ro (4) = (c(4)) = *(5)= 3, ror (5) (4) roc(6) r(+(6)) = -(3) = 1 Now = 2

le - not = 2 3 4 5 6 4 3 2 9 5 ii Want to calculate Zor. Now cor (1) = c(-6)) = {(5)= 4 Tor(2) = 2(+(2)) = < (6) = 3 zor(3) = + (-()) = 2(1) = 2 Tor (4) = {(-(4)) = {(2)= 1 zor(5) = (((s)) = < (3)=6 200(6) = {(+(6)) = {(4) = 5 le Tor 2 3 3 2 4 5 6 I 6 5

Want to calculate corol. = 20 (voz) Let = 6 201 Ae we have calculated ror. 1.e. g=rola 2 3 4 5 6 6 5 4 3 2 - and a= 2 6 - G + E E to 9 9 2 Now. thus To (roz) = cog tog() = (g(1)) = 266) = 3. tog(?) = (8(2)) = 2(5)= 4 Zog(3) = 7• (8(3)) = c(4)=5 Tog(4) = < (3{4)) = 2(3)=6 Zog(s) = c(g(5)) = + (3)=1 Tog(6) = 2(g(6)) = {(1)=2

therefore Zog - cogacor or 2 3 4 5 6 6 l 2 3 4 5 Sinice torot = 1 2 3 4 5 6 3 4 5 6 1 2 5 6 2 6 120 toroz 5 3 2 Here, We see, clearly is a rotation rotation with angle 120°, cock wise. Toror