cayley tables are stupid. it's much simpler to just list the rule by which a product is evaluated.

in this case:

a*b = ab (mod 35). note that a*b = b*a, since the integer ab = ba, so reducing either of these mod 35 gives the same result.

to find an identity (if there is one) we need to find an element e such that:

a*e = e*a = a, for all a in G. since e*a = a*e (see above), we only need to check for a right-identity.

for example 5 can't be the identity, since 5*5 = 25 ≠ 5 (take either 5 to be "a").

10 can't be the identity since 5*10 = 15 ≠ 5.

5*15 = 5, so it's possible 15 could be the identity. we could check the rest of the products (there's only 5 more to check), but let's work smart, not hard.

note that all of our elements are multiples of 5, so we can write them as 5x, x = 1,2,3,4,5,6.

now (5x)(15) = 75x = 70x + 5x.

hence (5x)(15) = (2x)(35) + 5x, so:

75x = 5x (mod 35).

this shows that (5x)*(15) = 5x for each x = 1,2,3,4,5,6, so that 15 is indeed our identity (why do case-by-case, when you can do all at once?).

now we turn to inverses (if we have them).

for every a, we need to find a b so that:

a*b = 15 (since * is commutative, this will mean b*a = 15).

well, we COULD examine our table, and find that 5^{-1}= 10, etc.

let's see if we can do this "algebraically".

suppose (5x)*(5y) = 15 (mod 35).

this means 25xy = 15 (mod 35).

note that this means:

25xy = 35k + 15, so we can divide by 5 to get:

5xy = 7k + 3, that is:

5xy = 3 (mod 7)

also: note that (3)(5) = 15 = 1 (mod 7) (because 15 = 2(7) + 1)

so if we multiply by 3, we get:

xy = 15xy = 9 = 2 (mod 7)

therefore y = 2x^{-1}(mod 7).

now we need to know the inverses of each x (mod 7). this is somewhat easier to compute:

1 has inverse 1 (duh!)

2 has inverse 4 (since 8 = 1(7) + 1)

3 has inverse 5 (we saw this above)

4 has inverse 2 (since 2 has inverse 4)

5 has inverse 3 (because....)

6 has inverse 6 (because that's the only inverse left...or: because 36 = 5(7) + 1, take your pick).

ok, let's put this back together for G:

5 = 5(1), and 1 has inverse 1 mod 7, and 2*1 = 2 (mod 7), so y = 2, so 5 has inverse 5(2) = 10.

10 = 5(2), and 2 has inverse 4 mod 7, and 2*4 = 8 = 1 (mod 7), so y = 1, so 10 has inverse 5.

15 should be its own inverse, but we'll check, just for laughs:

15 = 5(3), and 3 has inverse 5 mod 7, and 2*5 = 10 = 3 (mod 7), so y = 3, so 15 has inverse 15. hey, it works!

20 = 5(4), and 4 has inverse 2 mod 7, and 2*2 = 4 (mod 7), so y = 4, so 20 has inverse 20 (yes, a non-identity element CAN be its own inverse).

25 = 5(5), and 5 has inverse 3 mod 7, and 2*3 = 6 (mod 7), so y = 6, so 25 has inverse 30.

30 = 5(6), and 6 has inverse 6 mod 7, and 2*6 = 12 = 5 (mod 7), so y = 5, so 30 has inverse 25.

in fact, there's a reason why (mod 7) figured so prominently in this.

our group G is really "the same" (has the same cayley table) as the group {1,2,3,4,5,6} under multiplication mod 7:

15 <---> 1 = 3^{6}mod 7

25 <---> 2 = 3^{2}mod 7

5 <---> 3

30 <---> 4 = 3^{4}mod 7

10 <---> 5 = 3^{5}mod 7

20 <---> 6 = 3^{3}mod 7

what happens is that 3^{k}(mod 7) gets sent to 5^{k}(mod 35):

5^{2}= 25 mod 35

5^{3}= 125 = 20 mod 35

5^{4}= 625 = 30 mod 35

5^{5}= 3125 = 10 mod 35

5^{6}= 15625 = 15 mod 35