( 1)5? =25= 1, so |5|=2.

(2) Let k < G be a subgroup. Only discuss whether there are 5,7, 1 1:

(2. 1) Note 5? = 1,7? =49= 1, 1 1? = 12 1= 1, which are all second-order elements.

(2.2) If there are two other items in 5,7, 1 1, the third item can be generated, so

The subgroups are {1} {1, 5} {1, 7}, {1,1} and G, four of which are true subgroups.

2.

Are you sure it's 3H, not 3+H? This is an additive group, and the coset should be written as 3+H, unless N is used to induce modular 12 multiplication. ...

3+H={3+h|h∈H}={3，4， 1 1}，3H={3h|h∈H}={0， 12=0，24=0}={0}

3.

Bc can only be =a or b or c.

If bc=b or c, two sides are multiplied by the corresponding inverse element to get c= 1=a or b= 1=a, which is contradictory.

So bc=a, so C and B are reciprocal, so C and B are in the same order.

c？ Could be a, b, c

If c? =a is C? =a=bc deduces the contradiction of b = C.

If c? =c then cc=c deduces the contradiction of c = a.

So c? = b = c- 1, what about c? =a, so |c|=3

Similarly, b? =c

4. closed: any a, b ∈ z, a * b = a+b-2 ∈ z.

Yao Yuan: It's two o'clock.

The inverse of any a: it is 2-a.

So (z, *) is a group.

5. I know it directly through the counting formula (in abstract algebra, I don't know if you have studied it).

|G|=|K||G/K|

For group homomorphism f: g1->; G2, where | G 1 | = | section ||| IMF |

So for h: g->; G, where |G|=|K||h(G)|

So |K| is 4, |G/K|=|Im h|=|h(G)|=3.