For each of the following, either give an example or explain why no such example exists. One or two sentence answers will suffice; you don’t need to give formal proofs.
A non-abelian group all of whose proper subgroups are normal.
An abelian group with a subgroup which is not normal.
A non-abelian group of order \(p^2\) where \(p\) is prime.
An abelian group \(G\) of order \(n\) and a divisor \(d\) of \(n\) such that \(G\) has no subgroup of order \(d\).
An infinite group all of whose elements have finite order.
A prime \(p\) and a finite group \(G\) of order \(p^{k+1}m\) where \(p\) does not divide \(m\), such that \(G\) has non-isomorphic subgroups of order \(p^k\).
Solution
Write \(C_r=\mathbb Z/r\mathbb Z\).
1. In part (a), the quaternion group \(Q_8\) is an example.
The quaternion group is \(Q_8=\{1,-1,i,-i,j,-j,k,-k\}\), with \(i^2=j^2=k^2=-1\), \(ij=k\), and \(ji=-k\) [@DF04]. Thus it is nonabelian. By Lagrange’s theorem, a proper subgroup has order \(1\), \(2\), or \(4\). The identity subgroup is normal. The only element of order \(2\) is \(-1\), since each of \(\pm i,\pm j,\pm k\) squares to \(-1\). Hence the only subgroup of order \(2\) is \(\{1,-1\}\), which is central and normal. Every subgroup of order \(4\) has index \(2\) and is normal: for an element outside it, the left and right cosets are both its complement. This covers all proper subgroup orders.
2. No example exists in part (b).
If \(G\) is abelian and \(H\leq G\), then \(ghg^{-1}=h\) for every \(g\in G\) and \(h\in H\). Hence \(gHg^{-1}=H\) for every \(g\), so every subgroup is normal.
3. No example exists in part (c): every group of order\(p^2\) is abelian.
For a noncentral element, the size of its conjugacy class is \([G:C_G(g)]\), a power of \(p\) greater than \(1\), so it is divisible by \(p\). Splitting \(G\) into conjugacy classes therefore gives \(|G|\equiv|Z(G)|\pmod p\). The center contains the identity, and its order divides \(p^2\). Consequently \(|Z(G)|\) is \(p\) or \(p^2\).
If \(|Z(G)|=p^2\), then \(G=Z(G)\) is abelian. Otherwise \(G/Z(G)\) has prime order and is cyclic. Choose \(g\in G\) whose coset generates this quotient. Every element of \(G\) is \(g^a z\) for some integer \(a\) and \(z\in Z(G)\). Two such elements commute, because their powers of \(g\) commute and their central factors commute with everything. Thus \(G\) is abelian in this case as well.
4. No example exists in part (d).
We prove by induction on \(|G|\) that every finite abelian group has a subgroup of each positive order dividing \(|G|\). For \(d=1\), the identity subgroup works; this includes the base case \(|G|=1\).
Suppose \(d>1\) and choose a prime \(p\mid d\). Since \(p\mid|G|\), Cauchy’s theorem supplies an element of order \(p\), generating a subgroup \(N\) of order \(p\) [@DF04]. It is normal because \(G\) is abelian. The quotient \(G/N\) is abelian of order \(|G|/p<|G|\), and \(d/p\) divides its order. By induction it has a subgroup \(\overline H\) of order \(d/p\). The inverse image of \(\overline H\) in \(G\) is a subgroup consisting of \(d/p\) cosets of \(N\), each of size \(p\). Its order is \(d\), as required.
5. In part (e), take \(\bigoplus_{n\geq1}C_2\).
This is the group of sequences \((a_1,a_2,\ldots)\) with \(a_n\in\mathbb Z/2\mathbb Z\) and only finitely many nonzero coordinates, under coordinatewise addition. It is infinite because the sequences having a single \(1\) in position \(n\) are distinct as \(n\) varies. Every element added to itself is zero, so every element has order \(1\) or \(2\), and in particular finite order.
6. In part (f), take \(p=2\), \(k=2\), \(m=1\), and\(G=C_4\times C_2\).
This group has order \(8=2^{2+1}\cdot1\), with \(2\nmid1\). The subgroup \(H_1=\langle(1,0)\rangle\) has order \(4\) and is cyclic. The subgroup \[ H_2=\langle(2,0),(0,1)\rangle =\{(0,0),(2,0),(0,1),(2,1)\} \] also has order \(4=2^k\), but every nonidentity element has order \(2\). Thus \(H_2\cong C_2\times C_2\) and \(H_1\cong C_4\) are not isomorphic, since an isomorphism preserves element orders.