Let G be a group of order $|G|=24$. We need to determine which of the given statements are necessarily true for any such group G.

Group Properties and Sylow's Theorems

We can use Sylow's theorems to analyze the structure of G. The order of G is $24 = 2^3 \times 3^1$.

• The number of Sylow 2-subgroups, $n_2$, must satisfy $n_2 \equiv 1 \pmod{2}$ and $n_2$ must divide 24. Possible values for $n_2$ dividing 24 are 1, 2, 3, 4, 6, 8, 12, 24. The odd values are 1 and 3. So, $n_2 \in \{1, 3\}$.
• The number of Sylow 3-subgroups, $n_3$, must satisfy $n_3 \equiv 1 \pmod{3}$ and $n_3$ must divide 24. Possible values for $n_3$ dividing 24 are 1, 2, 3, 4, 6, 8, 12, 24. Values congruent to 1 mod 3 are 1 and 4. So, $n_3 \in \{1, 4\}$.

A Sylow p-subgroup is normal if and only if there is only one of them ($n_p = 1$).

Normal Subgroup Analysis

Let's evaluate the first statement:

Statement 1: G has a normal subgroup of order 3.

This statement is equivalent to saying that the number of Sylow 3-subgroups, $n_3$, is necessarily 1. However, $n_3$ can be either 1 or 4. For example, the symmetric group $S_4$ is a group of order 24. The Sylow 3-subgroups of $S_4$ are the cyclic groups generated by 3-cycles, like $\langle (123) \rangle$. There are $\binom{4}{3} \times (3-1)! / 3 = 4 \times 2 / 3$ (this formula is for counting cyclic subgroups, total 3-cycles are $\binom{4}{3} \times (3-1)! = 4 \times 2 = 8$). Each cyclic subgroup of order 3 contains 2 elements of order 3. So there are $8/2 = 4$ distinct subgroups of order 3 in $S_4$. Thus, for $S_4$, $n_3=4$. Since $n_3=4 > 1$, none of these subgroups are normal in $S_4$. Therefore, it is not necessarily true that a group of order 24 has a normal subgroup of order 3.

Simple Group Property

Let's evaluate the second statement:

Statement 2: G is not a simple group.

A simple group is a group whose only normal subgroups are the trivial group ({e}) and the group itself. For a group of order 24 to be simple, it cannot have any proper nontrivial normal subgroups (i.e., normal subgroups of order 2, 3, 4, 6, 8, 12). From Sylow's theorems, we know $n_2 \in \{1, 3\}$ and $n_3 \in \{1, 4\}$.

• If $n_2 = 1$, then the unique Sylow 2-subgroup of order 8 is normal in G. Since its order (8) is a proper divisor of 24 and greater than 1, this normal subgroup is proper and nontrivial. In this case, G is not simple.
• If $n_3 = 1$, then the unique Sylow 3-subgroup of order 3 is normal in G. Its order (3) is a proper divisor of 24 and greater than 1. In this case, G is not simple.

G can be simple only if $n_2 > 1$ AND $n_3 > 1$. This means we must have $n_2 = 3$ and $n_3 = 4$. Consider the case $n_2=3$. G acts on the set of its three Sylow 2-subgroups by conjugation. This action induces a group homomorphism $\phi: G \to S_3$, where $S_3$ is the symmetric group on 3 elements. The kernel of this homomorphism, $\text{ker}(\phi)$, is a normal subgroup of G. The image $\phi(G)$ is a subgroup of $S_3$. The order of $S_3$ is $3! = 6$. The order of the image $|\phi(G)|$ must divide $|S_3| = 6$. By the First Isomorphism Theorem, $|G/\text{ker}(\phi)| = |\phi(G)|$, so $24/|\text{ker}(\phi)| = |\phi(G)|$. Thus, $|\text{ker}(\phi)| = 24 / |\phi(G)|$. Since $|\phi(G)|$ divides 6, $|\phi(G)|$ can be 1, 2, 3, or 6. If $|\phi(G)| = 1$, then $\text{ker}(\phi) = G$. This means conjugation by any element fixes all Sylow 2-subgroups, implying they are all normal. But we assumed $n_2=3 > 1$, so they cannot all be normal. Thus $|\phi(G)| \neq 1$. Since $|\phi(G)|$ must be a divisor of 6 other than 1, it must be 2, 3, or 6. In all these cases, $|\text{ker}(\phi)| = 24/|\phi(G)|$ will be $24/2=12$, $24/3=8$, or $24/6=4$. In any case where $n_2=3$, the kernel of the homomorphism $\phi$ is a normal subgroup of order 4, 8, or 12. These are proper nontrivial normal subgroups. Therefore, regardless of the values of $n_2$ and $n_3$, G must have a proper nontrivial normal subgroup. This means G is necessarily not a simple group. Statement 2 is necessarily true.

Homomorphism Existence

Let's evaluate the third statement:

Statement 3: There exists an injective group homomorphism from G to S₈.

An injective group homomorphism from G to $S_8$ exists if and only if G is isomorphic to a subgroup of $S_8$. Cayley's Theorem states that any group G is isomorphic to a subgroup of the symmetric group $S_{|G|}$. For $|G|=24$, G is isomorphic to a subgroup of $S_{24}$. However, the statement is about $S_8$, not $S_{24}$. Consider the cyclic group $C_{24}$ of order 24. This is a group of order 24. $C_{24}$ contains an element of order 24. If $C_{24}$ was isomorphic to a subgroup of $S_8$, then $S_8$ must contain an element of order 24. The order of an element in a symmetric group $S_n$ is the least common multiple (lcm) of the lengths of the cycles in its disjoint cycle decomposition. For $S_8$, we need to find partitions of 8 and calculate the lcm of the parts. The maximum possible order of an element in $S_8$ is the largest lcm of a partition of 8. Partitions of 8 are, for example, 8 (lcm=8), 7+1 (lcm=7), 6+2 (lcm=6), 5+3 (lcm=15), 5+2+1 (lcm=10), 4+3+1 (lcm=12), etc. The maximum lcm for partitions of 8 is 15 (from 5+3). Since the maximum element order in $S_8$ is 15, $S_8$ does not contain an element of order 24. Therefore, $C_{24}$ cannot be isomorphic to a subgroup of $S_8$. Since $C_{24}$ is a group of order 24, it is not necessarily true that an injective group homomorphism exists from G to $S_8$ for *any* group G of order 24.

Subgroup Index

Let's evaluate the fourth statement:

Statement 4: G has a subgroup of index 4.

The index of a subgroup H in G is $[G:H] = |G|/|H|$. We are looking for a subgroup H such that $24/|H|=4$. This means $|H| = 24/4=6$. So the statement is equivalent to saying that G necessarily has a subgroup of order 6.

We consider the number of Sylow 3-subgroups, $n_3 \in \{1, 4\}$.

• If $n_3 = 1$, G has a unique, normal Sylow 3-subgroup, say P, with $|P|=3$. By Cauchy's Theorem, G must have an element of order 2, which generates a subgroup Q of order 2. Since P is normal, the product PQ is a subgroup of G. The order of PQ is $|P||Q|/|P \cap Q|$. Since $|P|=3$ and $|Q|=2$, their orders are coprime, so $P \cap Q = \{e\}$. Thus, $|PQ| = 3 \times 2 / 1 = 6$. So, if $n_3=1$, G has a subgroup of order 6.
• If $n_3 = 4$, G has four distinct Sylow 3-subgroups. Let P be any one of these Sylow 3-subgroups, so $|P|=3$. The normalizer of P in G, denoted by $N_G(P)$, is a subgroup of G. The index of the normalizer is related to the number of conjugates of P by the Orbit-Stabilizer Theorem: $[G:N_G(P)] = n_3$. Since $n_3=4$, we have $[G:N_G(P)] = 4$. The order of the normalizer is $|N_G(P)| = |G| / [G:N_G(P)] = 24 / 4 = 6$. Thus, if $n_3=4$, G has a subgroup of order 6 (namely, the normalizer of any Sylow 3-subgroup).

In both cases ($n_3=1$ and $n_3=4$), G must have a subgroup of order 6. A subgroup of order 6 has index $24/6=4$. Therefore, G necessarily has a subgroup of index 4. Statement 4 is necessarily true.

Based on the analysis, the statements that are necessarily true are Statement 2 and Statement 4.