Cosets

Theorem

Claim: \(g_1H=g_2H\) is possible even when \(g_1 \neq g_2\)

Proof: Consider a subgroup of \(S_3\) that only contains the rotations and the identity element \(e\). From above, this subgroup contains \((n-1)+1=n\) elements , so the subgroup, denoted by \(H\), has \(3\) elements, i.e. \(e, (123),(132)\). Now consider \((12) \in S_3\):

\[\begin{align} (12)(H)&=(12),(12)(123),(12)(132) \\ &=(12),(23),(13) \end{align}\]

Now consider \((23) \in S_3\):

\[\begin{align} (23)(H)&=(23),(23)(123),(23)(132) \\ &=(23),(13),(12) \\ &=(12),(23),(13) \\ &=(12)(H) \end{align}\]

Physical interpretation: It works for the rotation group of \(S_3\) because of symmetry, i.e. \(1,2,3\) can be relabelled between themselves freely as long as the relabelling itself is bijective

This shows that \(g_1H=g_2H\) is possible even when \(g_1 \neq g_2\) . The function \(f: G \rightarrow G\) s.t. \(g_n \rightarrow hg_n\) for any \(n \in \mathbb{N}\) is not injective as \((f(g_1)=f(g_2)) \nrightarrow (g_1=g_2)\) \(\blacksquare\)

Surprising Theorem: \(\forall h \in H, hH=H\)

Idea of proof: From set theory if \(hH \subseteq H\) and \(H \subseteq hH\) then \(hH=H\).

\(hH\subseteq H\) is obvious from group axioms (closure). I’ll prove \(H \subseteq hH\). To prove that I’ll need to show \(y \in H \rightarrow y \in hH\).

Working backwards, we want \(h^{-1}y \in H\), This can easily be proven true: If \(h \in H\) then \(h^{-1} \in H\) by group axioms (inverse). Then if \(y \in H\) and \(h^{-1} \in H\), \(h^{-1}y \in H\) by group axioms (closure).

More formal approach: Pick \(y \in H\). Since \(h \in H\), \(h^{-1} \in H\) by inverse axiom of groups, hence \(h^{-1}y \in H\) by closure property of groups. Denote \(m=h^{-1}y\). As \(m \in H\), \(hm \in hH\), hence \(hh^{-1}y \in hH\), i.e. \(y \in hH\). Hence \(H \subseteq hH\).

As \(hH \subseteq H\) and \(H \subseteq hH\) then \(hH=H\) \(\blacksquare\)

Example: As above, the rotation group of \(S_3\) is a subgroup of \(S_3\) (Left as an exercise for the readers). Pick \(h=(123)\):

\[\begin{align} (123)(H)&=(123),(123)(123),(123)(132) \\ &=(123),(132),e \\ \end{align}\]

From above, the rotation group is re-obtained.

Claim:

Suppose that \(H \leq G\) and \(H\) is finite. Then \(|g H|=|H|\) for all \(g \in G\).

Proof:

Consider the map \(H \rightarrow g H\) given by \(h \mapsto g h\). We would like to prove it is a bijection. To prove that it is a bijection, we need to prove it is both injective and surjective.

It is injective since \(g h_1=g h_2\) implies that \(h_1=h_2\) (From Lemma above)

Surjection is obvious from definition (I’ll leave it as an exercise for the readers)

\(\Rightarrow\) \(H \rightarrow g H\) is a bijection \(\Rightarrow\) \(|g H|=|H| \quad \forall g \in G\) \(\blacksquare\)

Converse of Lagrange’s Theorem

Converse of Lagrange’s Theorem is not generally true , i.e. if \(d\) is a divisor of the order of a group \(G\), then there may not exist a subgroup \(H\) where \(H = d\)

(I can’t type absolute signs)

Proof

An example: \(A_4\) has order \(12\) but doesn’t have subgroups of order \(6\).

The proof below is not original and comes from here

Assume \(H < A_4\) is a subgroup of \(A_4\) of order 6

Then, for any \(a \notin H\), \(a H \cap H=\emptyset\)

Again, since all cosets have name number of elements,\(a H=e H=H\) (all cosets have the same number of elements), this implies that \(a H=6\).

(I can’t type absolute signs)

Then, as cosets form a partition of the group \(A_4\), and \(\left|A_4\right|=12\), then \(A_4=H \cup a H\)

Now suppose that \(a\) is a 3-cycle in \(A_4\), then either \(a^2 \in H\) or \(a^2 \in a H\) (by closure if \(a \in A_4\) then \(a^2 \in A_4\))

If \(a^2 \in H\), then this implies that \(a^4=a^2 \cdot a^2 \in H\) (again, closure) However, since the order of \(a\) is 3 (it is a 3-cycle), then \(a^4=a\) and hence \(a \in H\). However, from assumption \(a \notin H\). Contradiction

Similarly, if \(a^2 \in a H\) then \(a \in H\) and again this is a contradiction. Hence, \(H\) cannot be a subgroup of \(A_4\) of order 6 \(\blacksquare\)

We have assumed implicitly that \(A_4\) is generated by 3-cycles: I won’t proof that but I’ll give an idea on why this might be true:

First, any product of two transpositions is a product of 3-cycles.

As an example, consider the product of two transpositions:

\((ab)(ac)=(acb)\) \((ab)(cd)=(dac)(abd)\)

Hint: Do the “inside” operation first, e.g. for the first one consider \((ac)\) then consider \((ab)\). In general, for \((f \circ g)(x)=f(g(x))\)

I’ll do the first one as an example.

To see where \(a\) is mapped to, First consider \((ac)\), \(a \rightarrow c\). There is no \(c\) in \((ab)\) so indeed \(a \rightarrow c\).

To see where \(c\) is mapped to, First consider \(c \rightarrow a\) in \((ac)\). Notice that \(a \rightarrow b\) in \((ab)\). So \(c \rightarrow b\).

Finally for \(b\), again, first consider \((ac)\). However, \(b\) is not in \((ac)\). From \((ab)\), we can see \(b \rightarrow a\). Thus, \(b \rightarrow a\)

Therefore, \(a \rightarrow c \rightarrow b \rightarrow a\), i.e. \((ab)(ac)=(acb)\) \(\blacksquare\)

I will state without proof that the product of a even number of transpositions is also a 3-cycle.