If G is a group of order 12, the G must have a subgroup of all of the following orders EXCEPT
Choose one answer.
a. 0
b. 12
c. 2
d. 3
e. 6
f. 4
if H is a subgroup of G, |H| divides |G|
so, (a)
To determine which of the given orders can definitely be a subgroup of a group of order 12, we can use Lagrange's Theorem.
Lagrange's Theorem states that the order of any subgroup of a finite group divides the order of the group. In other words, if G is a group of order n and H is a subgroup of G with order m, then m divides n.
In this case, since G has order 12, any subgroup of G must have an order that divides 12.
Now let's consider each option:
a. 0: The trivial subgroup, consisting of just the identity element, always exists in any group. Therefore, a subgroup of order 0 is possible.
b. 12: Since G itself has order 12, it is also a subgroup of itself. Therefore, a subgroup of order 12 is possible.
c. 2: The order 2 divides 12, so it is possible to have a subgroup of order 2.
d. 3: The order 3 also divides 12, so it is possible to have a subgroup of order 3.
e. 6: The order 6 also divides 12, so it is possible to have a subgroup of order 6.
f. 4: The order 4 divides 12, so it is possible to have a subgroup of order 4.
So, the answer is option a. 0. A subgroup of order 0 is not possible because a subgroup must contain at least the identity element.