Subgroups, cosets, Lagrange, cyclic groups

Which of the following statements are true?

A.Let $G_{1}$ and $G_{2}$ be finite groups such that | $G_{1}$ | and | $G_{2}$ | are coprime. Then any homomorphism from $G_{1}$ to $G_{2}$ is trivial.✓ Correct
B.Let G be a finite group and f : G → G a homomorphism that fixes more than half of the elements of G. Then f(x) = x for all x ∈ G.✓ Correct
C.Let G be a finite group having exactly 3 subgroups. Then G is of order $p^{2}$ for some prime p.✓ Correct
D.Any finite abelian group G has at least d(|G|) subgroups, where d(m) is the number of positive divisors of m.✓ Correct

The idea, in plain words

Each of these is a counting argument in disguise. The size of an image has to divide two different totals, a set of fixed points is itself a subgroup so its size divides the whole, and having exactly three subgroups is a strong enough restriction to pin the size down completely.

Solution

(1) the image has order dividing both.

(2) the fixed set is a subgroup of index < 2

hence all of G.

(3) exactly three subgroups forces cyclic of order $p^{2}$ .

(4) an abelian group has a subgroup of every order dividing |G|.

Why each option is right or wrong

Checked against 1 symbolic computation

A.Correct
True.

The image $f(G_{1})$ is a subgroup of $G_{2}$

so its order divides | $G_{2}$ |; it is also isomorphic to $G_{1}/\ker f$

so its order divides | $G_{1}$ |.

A number dividing two coprime integers is 1

so the image is trivial.

The trigger: 'coprime orders' plus 'homomorphism' means push the image through both divisibility constraints.
B.Correct
True.

The fixed set {x : f(x) = x} is a subgroup of G — it contains the identity and is closed under products and inverses.

By Lagrange its order divides |G|, and it exceeds |G|/2

so its index is strictly less than 2.

The only positive integer below 2 is 1

so the fixed set is all of G.

'More than half' works because there is no room for an index-2 subgroup.
C.Correct
True.

Exactly three subgroups means {e}, G, and precisely one proper non-trivial subgroup.

Such a G must be cyclic (any non-generator generates a proper subgroup, and two distinct ones would give a fourth subgroup), and a cyclic group of order n has one subgroup per divisor of n — so n must have exactly three divisors

forcing $n = p^{2}$ .

For contrast, SymPy gives divisor_count(12) = 6

so $\mathbb{Z}/12$ has six subgroups.
D.Correct
True.

A finite abelian group has a subgroup of every order dividing |G| — the converse of Lagrange

which holds for abelian groups though not in general.

That gives at least one subgroup per divisor

hence at least d(|G|).

All four statements here are correct.

Last revised 11 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

Subgroups, cosets, Lagrange, cyclic groups

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