September 17th, 2026

Updated 19 Sep 2026Notes (PDF)

Recall that, for a subgroup $H \subseteq G$ , the left cosets $\{aH: a \in G\}$ of $H$ form a partition $G$ , since it is the set of equivalence classes from the equivalence relation $x \sim y$ if and only if $x, y \in G$ belong to the same left coset.

[0.0.30]Theorem(Cosets are the Same Order as Subgroup)#

Suppose $H$ is some subgroup of $G$ . Then $|aH| = |Ha| = |H|$ for every $a \in G$ .

Proof.

Fix some $a \in G$ . Then consider the map $f: aH \to H$ given by $f(x) = a^{-1}x$ . For each $h \in H$ , note that $f(ah) = a^{-1}(ah) = h$ , and so $f$ is surjective. Moreover, if $h_1, h_2 \in aH$ are such that $f(h_1) = f(h_2)$ , then note

a^{-1}h_1 = a^{-1}h_2 \implies h_1 = h_2,

and so $f$ is injective. Thus $f$ is bijective, and so $|aH| = |H|$ as desired.

An analogous procedure can be performed to show $g: Ha \to H$ given by $g(x) = xa^{-1}$ is also a bijection.

From now on, note that using the number of left cosets is interchangeable with using the number of right cosets.

[0.0.31]Definition(Index of a Subgroup)#

Let $H$ be a subgroup of $G$ . Then the index of $H$ , denoted $[G:H]$ , is the number of left cosets $H$ has.

Recall that the left cosets of a subgroup $H$ partition the group $G$ , and that the left cosets of $H$ all have the same order, which is $|H|$ . This lets us relate the order of $G$ with the cosets of $H$ .

[0.0.32]Proposition(Counting Formula)#

Let $H$ be a subgroup of $G$ . Then $|G| = [G:H] \times |H|$ .

This theorem reads that the order of a group $G$ is the number of cosets a subgroup has, multiplied by the size of each coset, which is obviously true. Since $[G:H]$ is an integer, we note that $|H|$ is an integer multiple of $|G|$

[0.0.33]Theorem(Lagrange's Theorem)#

Let $G$ be a group. Then, for any subgroup $H \subseteq G$ , we have that $|H|$ divides $|G|$ .

Lagrange's theorem gives rise to multiple important structure theorems about groups.

[0.0.34]Theorem(Element Order Divides Group Order)#

Let $G$ be a group. For every $g \in G$ , note $|g|$ divides $|G|$ .

Proof.

Note $\angled{g}$ is a subgroup of $G$ of order $g$ . The result follows by Lagrange.

[0.0.35]Theorem(Groups of Prime Order are Cyclic)#

Suppose $G$ is a group such that $|G| = p$ for some prime $p$ . Then $G \cong C_p$ for all cyclic groups of order $p$ .

Proof.

$G$ is not of order $1$ , since $1$ is not prime. Thus there must be some $g \in G$ such that $g \ne \id_G$ . Notably, since $|g|$ divides $|G| = p$ , we either have that $|g| = 1$ or $|g| = p$ . Note that $|g| \ne 1$ since $g \ne \id_G$ , and so $|g| = p$ . Thus $\angled{g}$ is a subgroup of $G$ with the same order as $G$ , and so $\angled{g} = G$ , completing the proof.

Note that all cyclic groups of order $n$ are isomorphic to one another. Indeed, for any cyclic group $G = \angled{g}$ of order $n$ , the map $\phi: \bZ/n\bZ \to G$ defined by $\phi(k) = g^k$ forms an explicit isomorphism. Since every cyclic group of order $n$ is isomorphic to $\bZ/p\bZ$ , they are all isomorphic to one another by transitivity.

[0.0.36]Theorem(Counting Formula for Homomorphisms)#

Let $\phi: G \to H$ be a homomorphism of finite groups. Then $|G| = |\ker\phi| \times |\im\phi|$ .

Proof.

Recall from earlier that there are exactly $|\im \phi|$ number of cosets of $\ker(\phi)$ that partition $G$ , each of which has order $|\ker(\phi)|$ .

[0.0.37]Definition(Normal Subgroup)#

Let $H$ be a subgroup of $G$ . Then $H$ is said to be a normal subgroup if, for all $g \in G$ and $h \in H$ , we have that $ghg^{-1} \in H$ . That is, a subgroup is normal if conjugation of its elements is a closed operation.

[0.0.38]Theorem(Equivalent Characterizations of Normal Subgroups)#

Let $G$ be a group and $H$ a subgroup. Then the following statements are equivalent.

For every $g \in G$ and $h \in H$ , $ghg^{-1} \in H$ .
For every $g \in G$ , $gHg^{-1} \subseteq H$ , where $gHg^{-1} = \{ghg^{-1}: h \in H\}$ .
For every $g \in G$ , $gHg^{-1} = H$ .
For every $g \in G$ , $gH = Hg$ .

Proof.

$(1) \implies (2)$ : Fix $g \in G$ . Since $ghg^{-1} \in H$ for every $h \in H$ , note that $gHg^{-1} \subseteq H$ as desired.

$(2) \implies (3)$ : Note that $gHg^{-1} \subseteq H$ for all $g \in G$ , so we seek to show $H \subseteq gHg^{-1}$ . Note that $g^{-1}Hg \subseteq H$ as well, and so conjugation of both sides yields

g(g^{-1}Hg)g^{-1} \subseteq gHg^{-1} \implies H \subseteq gHg^{-1},

and so $gHg^{-1} = H$ .

$(3) \implies (4)$ : Suppose $gHg^{-1} = H$ for all $g \in G$ . Right multiplying both sides by $g$ yields $gH = Hg$ .

$(4) \implies (1)$ : Suppose $gH = Hg$ for every $g \in G$ . Fix $g \in G$ . For any $h \in H$ , there exists some $h' \in H$ such that $gh = h'g$ . Then note $ghg^{-1} = h' \in H$ for every $h \in H$ . Since $g$ was arbitrary, this holds for every $g \in G$ , completing the proof.

Now we show some general groups we work with are normal subgroups.

[0.0.39]Theorem(Subgroups of Abelian Groups are Normal)#

Suppose $G$ is abelian. Then all subgroups $H$ of $G$ are normal.

Proof.

Fix any subgroup $H \subseteq G$ and $g \in G$ . Then note $ghg^{-1} = gg^{-1}h = h \in H$ for every $h \in H$ , and so $H$ is normal.

[0.0.40]Theorem(Subgroups of Index

2

are Normal)#

Suppose $G$ is a group and $H$ is a subgroup such that $[G:H] = 2$ . Then $H$ is a normal subgroup.

Proof.

Since $[G:H] = 2$ , note that $H$ partitions $G$ into exactly two left cosets and exactly two right cosets. Because $H$ itself is always a coset, the two left cosets must be $H$ and its complement $G \setminus H$ . Symmetrically, the two right cosets must also be $H$ and $G \setminus H$ .

We show that $gH = Hg$ for any $g \in G$ . If $g \in H$ , then $gH = H = Hg$ immediately. If $g \in G \setminus H$ , then $gH \ne H$ . Since left cosets partition $G$ , $gH$ must be the only other available left coset, namely $G \setminus H$ . Symmetrically, $Hg \ne H$ , meaning $Hg$ must be the only other available right coset, which is also $G \setminus H$ . Thus $gH = G \setminus H = Hg$ . In all cases, $gH = Hg$ , proving $H$ is normal.

[0.0.41]Definition(Center of Group)#

Let $G$ be a group. The center of $G$ , denoted $Z(G)$ , is the set of elements in $G$ that commute with every $g \in G$ . That is,

Z(G) = \{z \in G: zg = gz \text{ for all } g \in G\}.

[0.0.42]Theorem(Center is a Normal Subgroup)#

Let $G$ be a group. Then $Z(G)$ is a normal subgroup.

Proof.

Fix $g \in G$ , and let $h \in Z(G)$ be arbitrary. Then note $ghg^{-1} = gg^{-1}h = h \in Z(G)$ , and so $Z(G)$ is normal.

[0.0.43]Theorem(Kernel of Homomorphism is Normal Subgroup)#

Let $\phi: G \to H$ be a homomorphism. Then $\ker(\phi)$ is a normal subgroup of $G$ .

Proof.

Fix $g \in G$ . Suppose $h \in \ker(\phi)$ . Then note

\phi(ghg^{-1}) = \phi(g) \phi(h) \phi(g^{-1}) = \phi(g)\phi(g^{-1}) = \phi(\id_G) = \id_H,

and so $ghg^{-1} \in \ker(\phi)$ as desired.

[0.0.44]Theorem(Pushforward and Pullback on Normal Subgroups)#

Let $\phi: G \to G'$ be a homomorphism. Let $H$ be a subgroup of $G$ and $H'$ a subgroup of $G'$ .

$\phi^{-1}(H')$ is a normal subgroup of $G$ .
$\phi(H)$ is a normal subgroup of $G'$ if $\phi$ is surjective.

Proof.

Fix $g \in G$ , and let $\phi^{-1}(H')$ . We show that, for any $k \in \phi^{-1}(H)$ , we also have that $gkg^{-1} \in \phi^{-1}(H)$ , making $\phi^{-1}(H)$ normal. Note that $H'$ is normal, and so $\phi(g)\phi(k)[\phi(g)]^{-1} \in H'$ by definition. Thus $\phi(gkg^{-1}) \in H'$ , and so $gkg^{-1} \in \phi^{-1}(H')$ as desired.
Let $y \in \phi(H)$ and $g' \in G'$ . We seek to show that $g'y(g')^{-1} \in \phi(H)$ . Because $y \in \phi(H)$ , there is some $h \in H$ such that $\phi(h) = y$ . Furthermore, because $\phi$ is surjective, there is some $g \in G$ such that $\phi(g) = g'$ . Substituting these yields
$g'y(g')^{-1} = \phi(g)\phi(h)[\phi(g)]^{-1} = \phi(ghg^{-1}).$
Since $H$ is a normal subgroup of $G$ , we have that $ghg^{-1} \in H$ . Thus its image $\phi(ghg^{-1})$ is strictly contained in $\phi(H)$ , meaning $g'y(g')^{-1} \in \phi(H)$ . Thus $\phi(H)$ is normal in $G'$ .