Criteria for Normal Subgroups

Method 1:

Let $H$ be a subgroup of group $G$. Then $H \lhd G \iff \forall g \in G, \ gH = Hg \ (\text{or} \ gHg^{-1} = H)$.

Method 2:

Let $H$ be a subgroup of group $G$. Then $H \lhd G \iff \forall g \in G, h \in H, \ ghg^{-1} \in H$.

Method 3:

Let $H$ be a subgroup of group $G$, and suppose that $[G : H] = 2$. Then $H \lhd G$.

Proof: For $\forall g \in G$, if $g \in H$, then $gH = H = Hg$. If $g \notin H$, then $G = H \cup gH = H \cup Hg$. Hence, it must be that $gH = Hg$. Therefore, for $\forall g \in G$, $gH = Hg$, so $H \lhd G$.

Method 4:

Let $H$ be a subgroup of a finite group $G$, and suppose that $|H| = m$, and there is only one subgroup of order $m$ in $G$. Then $H \lhd G$.

Proof: Since $\forall g \in G, |gHg^{-1}| = |H| = m$, and since there is only one subgroup of order $m$ in $G$, we must have $gHg^{-1} = H$. Therefore, $gH = Hg$, and thus $H \lhd G$.

Method 5:

Let $H$ be a subgroup of group $G$, and define $N(H) = \{ g \in G \mid gH = Hg \}$, which is called the normalizer of $H$ in $G$. Then $H \lhd G$ if and only if $N(H) = G$.

Method 6:

Let $H$ be a subgroup of a finite group $G$, and suppose $[G : H]$ is a prime number. If there exists an element $g \in G$ such that $g \notin H$ and $gH = Hg$, then $H \lhd G$.

Proof: Let $|G| = n$, $|H| = m$, and $[G : H] = p$. Consider $N(H)$, the normalizer of $H$ in $G$. It is easy to prove that $H \lhd N(H) \le G$. Let $|N(H)| = t$. By Lagrange’s theorem, we have:

$$
m = np \tag{1}
$$

$$
n | t,\quad t | m
$$

Hence, there exist $s, k \in \mathbb{Z}$ such that:

$$
t = ns \tag{2}
$$

$$
m = tk \tag{3}
$$

Combining (1), (2), and (3), we get:

$$
np = m = nsk
$$
which implies:

$$
sk = p
$$

Since $p$ is prime and $H$ is a proper subgroup of $N(H)$, we know $s > 1$ and $n < t$. Therefore, we must have $s = p$ and $k = 1$. Thus, $t = np = m$, and therefore $N(H) = G$, so $H \lhd G$.

Method 7:

Let $G$ be a group, and suppose $H$ is the kernel of some homomorphism from $G$. Then $H \lhd G$.

Method 8:

Let $H$ be a subgroup of group $G$. If $H = \text{Core}_G(H)$, then $H \lhd G$.

The cover image of this article was taken in Singapore.