Groups

Prove or disprove: If \(H\) and \(K\) are subgroups of a group \(G\), then \(HK = \{hk: h \in H \text{ and } k \in K\}\) is a subgroup of \(G\). What if \(G\) is abelian?

\(Proof.\) Suppose \(HK \leqslant G\), then \(\forall x \in HK\), \(x^{-1} \in HK\) too, suppose \(x^{-1} = hk\), so \(x = (hk)^{-1} = k^{-1}h^{-1} \in KH\), thus \(HK \subseteq KH\). \(\forall x = kh \in KH\), \(x^{-1} = h^{-1}k^{-1} \in HK\), so \(x = x^{-1^{-1}} \in HK\), thus \(KH \subseteq HK\). Hence \(HK = KH\).

Not all subgroups \(H\) and \(K\) satisfy the condition \(HK = KH\), so we can't say that \(HK\) must be a subgroup of \(G\).

But if \(HK = KH\), \(\forall hk\) and \(h'k' \in HK\), \[hk(h'k')^{-1} = hk{k'}^{-1}{h'}^{-1} = hk''{h'}^{-1} = hh''k''' = h'''k''' \in HK\], so \(HK \leqslant G\).

Hence we can get the final conclusion: \(HK \leqslant G\) if and only if \(HK = KH\). \(\blacksquare\)

Prove that \(\text{det}(AB) = \text{det}(A)\text{det}(B)\) in \(GL_2(\mathbb{R})\). Use this result to show that the binary operation in the group \(GL_2(\mathbb{R})\) is closed; that is, if \(A\) and \(B\) are in \(GL_2(\mathbb{R})\), then \(AB \in GL_2(\mathbb{R})\).

\(Proof.\) \[ {\begin{vmatrix} A \end{vmatrix}} {\begin{vmatrix} B \end{vmatrix}} = {\begin{vmatrix} A & 0 \\ -I & B \end{vmatrix}} = {\begin{vmatrix} 0 & AB\\ -I & B \end{vmatrix}} = {\begin{vmatrix} AB \end{vmatrix}} \] hence \(\text{det}(AB) = \text{det}(A)\text{det}(B)\). \(\forall A, B \in GL_2(\mathrm(R))\), \(\text{det}(AB) = \text{det}(A)\text{det}(B)\), so \(AB\) is an invertible matrix, thus \(AB \in GL_2(\mathrm(R))\). \(\blacksquare\)

Prove or disprove that every group containing six elements is abelian.

\(Solution.\) \(D_3\) has six elements but is not abelian. \(\blacksquare\)

Give a specific example of some group \(G\) and elements \(g, h \in G\) where \((gh)^n \not = g^nh^n\).

\(Solution.\) Suppose \(G = S_3\), \(g = {\begin{pmatrix} A & B & C\\ B & C & A \end{pmatrix}}, h = {\begin{pmatrix} A & B & C\\ A & C & B \end{pmatrix}}, n = 2\), then \[ (gh)^2 = ({\begin{pmatrix} A & B & C\\ B & C & A \end{pmatrix}} {\begin{pmatrix} A & B & C\\ A & C & B \end{pmatrix}})^2 ={\begin{pmatrix} A & B & C\\ B & A & C \end{pmatrix}}^2 ={\begin{pmatrix} A & B & C\\ A & B & C \end{pmatrix}} \], but \[ g^2h^2 = {\begin{pmatrix} A & B & C\\ B & C & A \end{pmatrix}}^2 {\begin{pmatrix} A & B & C\\ A & C & B \end{pmatrix}}^2 = {\begin{pmatrix} A & B & C\\ C & A & B \end{pmatrix}} \not = (gh)^2 \qed \]

Prove or disprove: If \(H\) and \(K\) are subgroups of a group \(G\), then \(HK = \{hk: h \in H \text{ and } k \in K\}\) is a subgroup of \(G\). What if \(G\) is abelian?

\(Solution.\) If \(H \subseteq K\) or \(K \subseteq H\), it's obvious that \(HK \leqslant G\).

If \(\exists h \text{ and } k \in G\), \(h \in H, h \not\in K, k \in K, k \not\in H\), then \(h \in H \cup K\), \(k \in H \cup K\), but \(hk \not \in H\), \(hk \not \in K\), so \(hk \not \in H \cup K\), thus \(H \cup K\) is not a subgroup of \(G\). The condition of G is abelian won't change the result. \(\blacksquare\)