Show that the intersection of two subgroups is again a subgroup.
- Hint: one-step subgroup test.
Show that if \(H\cong C_m\) and \(K\cong C_n\) are cyclic subgroups of \(G\), then \(|H\intersect K|\) divides \(\gcd(m,n)\). If \(H\) and \(K\) lie in a common cyclic subgroup of \(G\), show that \(H\intersect K\cong C_d\) where \(d=\gcd(m,n)\).
Show that the intersection of two subgroups with coprime orders is trivial.
Show that the union of two subgroups \(H,K\) is a subgroup iff \(H\subseteq K\) or \(K\subseteq H\), and so is generally not a subgroup.
Show that subgroups with the same prime order are either equal or intersect trivially.
Important for Sylow theory: show (perhaps by example) that if \(S_1, S_2\) are distinct subgroups of order \(p^k\), then it’s possible for their intersection to be trivial or for them to intersect in a subgroup of order \(p^\ell\) for \(1\leq \ell \leq k-1\).
Give a counterexample where \(H,K\leq G\) but \(HK\) is not a subgroup of \(G\).
Solution
1. The intersection of two subgroups is a subgroup.
Let \(H,K\le G\). The identity lies in both subgroups, so \(H\cap K\) is nonempty. If \(x,y\in H\cap K\), then \(x,y\in H\) and \(x,y\in K\). Since both are subgroups, \[ xy^{-1}\in H \qquad\text{and}\qquad xy^{-1}\in K. \] Hence \(xy^{-1}\in H\cap K\). The one-step subgroup test gives \(H\cap K\le G\).
2. If \(H\cong C_m\) and \(K\cong C_n\), then\[ |H\cap K|\mid\gcd(m,n). \]If moreover \(H\) and \(K\) lie in a common cyclic subgroup, then\[ H\cap K\cong C_{\gcd(m,n)}. \]
By 1, \(H\cap K\) is a subgroup of both \(H\) and \(K\). Lagrange’s theorem therefore gives \[ |H\cap K|\mid m, \qquad |H\cap K|\mid n, \] so \(|H\cap K|\mid\gcd(m,n)\).
Now suppose \(H,K\le C\) for a cyclic group \(C\). Put \(d=\gcd(m,n)\). A cyclic group has a unique subgroup of each order dividing its order. Let \(D\le C\) be the unique subgroup of order \(d\). Since \(d\mid m\) and \(d\mid n\), one has \(D\le H\) and \(D\le K\), hence \(D\le H\cap K\). The first paragraph gives \(|H\cap K|\le d\), while \(|D|=d\), so \(H\cap K=D\cong C_d\).
3. Subgroups of coprime finite orders intersect trivially.
If \(|H|\) and \(|K|\) are coprime, then by 2 the order of \(H\cap K\) divides both \(|H|\) and \(|K|\). Hence \[ |H\cap K|=1, \] so \(H\cap K=\{e\}\).
4. The union \(H\cup K\) is a subgroup if and only if \(H\subseteq K\) or \(K\subseteq H\).
If \(H\subseteq K\), then \(H\cup K=K\); similarly if \(K\subseteq H\).
Conversely, suppose \(H\cup K\) is a subgroup and neither subgroup contains the other. Choose \[ h\in H\setminus K, \qquad k\in K\setminus H. \] Since \(H\cup K\) is assumed to be a subgroup, \(hk\in H\cup K\). If \(hk\in H\), then \[ k=h^{-1}(hk)\in H, \] a contradiction. If \(hk\in K\), then \[ h=(hk)k^{-1}\in K, \] again a contradiction. Thus one subgroup must contain the other.
5. Two subgroups of the same prime order are either equal or intersect trivially.
Let \(|H|=|K|=p\) with \(p\) prime. By Lagrange, \[ |H\cap K|\mid p, \] so \(|H\cap K|\) is either \(1\) or \(p\). In the first case the intersection is trivial. In the second case \(H\cap K\) has the same order as both \(H\) and \(K\), hence \[ H=H\cap K=K. \]
6. Distinct subgroups of order \(p^k\) can have trivial intersection or intersection of order \(p^\ell\) for every \(1\le\ell\le k-1\).
Work in elementary abelian \(p\)-groups, viewed as vector spaces over \(\FF_p\).
For trivial intersection, take \[ G=\FF_p^{2k}, \quad S_1=\spanof\{e_1,\ldots,e_k\}, \quad S_2=\spanof\{e_{k+1},\ldots,e_{2k}\}. \] Then \(|S_1|=|S_2|=p^k\) and \(S_1\cap S_2=0\).
For \(1\le\ell\le k-1\), take \[ G=\FF_p^{2k-\ell}, \] and define \[ S_1=\spanof\{e_1,\ldots,e_k\}, \] \[ S_2=\spanof\{e_1,\ldots,e_\ell,e_{k+1},\ldots,e_{2k-\ell}\}. \] Both have dimension \(k\), hence order \(p^k\), while \[ S_1\cap S_2=\spanof\{e_1,\ldots,e_\ell\} \] has dimension \(\ell\) and order \(p^\ell\).
7. A product \(HK\) of subgroups need not be a subgroup.
Take \(G=S_3\), \[ H=\langle(12)\rangle, \qquad K=\langle(23)\rangle. \] Then \[ HK=\{e,(12),(23),(123)\}, \] so \(|HK|=4\). If \(HK\) were a subgroup of \(S_3\), Lagrange’s theorem would force \(4\mid6\), which is false. Hence \(HK\) is not a subgroup.