Let \(G\) be a nonabelian group of order \(16\) containing a subgroup \[ H\cong C_2\times C_2\times C_2. \] Prove that the number of elements of \(G\) of exact order \(2\) is either \(7\) or \(11\).
Solution
1. The subgroup \(H\) is normal in \(G\), and it contains exactly sevenelements of order \(2\).
Since \(\abs{G}=16\) and \(\abs{H}=8\), the subgroup \(H\) has index \(2\) in \(G\) and is therefore normal. Moreover, \[ H\cong C_2\times C_2\times C_2 \] has eight elements, and every nonidentity element has order \(2\). Hence \(H\) contains exactly seven elements of order \(2\).
2. If \(G\setminus H\) contains no element of order \(2\), then \(G\) hasexactly \(\boxed{7}\) elements of order \(2\).
By step 1, the seven nonidentity elements of \(H\) are all the involutions in this case.
3. Suppose that \(G\setminus H\) contains an element \(d\) of order \(2\).Then conjugation by \(d\) induces a nonidentity involution\[ D:H\longrightarrow H. \]
Normality of \(H\) from step 1 shows that conjugation by \(d\) preserves \(H\). Since \(d^2=e\), the induced automorphism \(D\) satisfies \(D^2=I\). Also \(G=H\sqcup dH\), so \(G\) is generated by \(H\) and \(d\). If \(D=I\), then \(d\) commutes with every element of the abelian group \(H\), which would make \(G\) abelian. This contradicts the hypothesis. Thus \(D\ne I\).
4. Regard \(H\) as the three-dimensional vector space \(\FF_2^3\).Then \(D\) is \(\FF_2\)-linear and\[ \dim_{\FF_2}\ker(D-I)=2, \]so \(D\) fixes exactly four elements of \(H\).
Every group homomorphism between elementary abelian \(2\)-groups is \(\FF_2\)-linear, so step 3 gives \(D\in\operatorname{GL}(H)\). Set \(N=D-I\). Because \(D^2=I\) and the ground field has characteristic \(2\), \[ N^2=(D-I)^2=D^2-2D+I=0. \] Hence \(\operatorname{im}N\subseteq\ker N\). If \(r=\operatorname{rank}N\), rank-nullity gives \[ r\le 3-r, \] so \(r\le1\). Step 3 gives \(N\ne0\), hence \(r=1\). Therefore \(\dim\ker N=3-1=2\), and the fixed space has \(2^2=4\) elements.
5. In the situation of step 3, the coset \(G\setminus H=dH\)contains exactly four elements of order \(2\).
Every element of \(dH\) has the form \(dh\) with \(h\in H\). Since \(d^2=e\), \[ (dh)^2=(dhd^{-1})h=D(h)h. \] Every element of \(H\) is its own inverse, so \((dh)^2=e\) exactly when \(D(h)=h\). By step 4 there are exactly four such \(h\). Since \(dh\notin H\), none of these four elements is the identity, and all four have exact order \(2\).
6. If \(G\setminus H\) contains an element of order \(2\), then \(G\) hasexactly \(\boxed{11}\) elements of order \(2\).
Step 1 gives seven involutions in \(H\), and step 5 gives four more in the disjoint coset \(G\setminus H\). Thus the total is \(7+4=11\).
7. Q.E.D.
Steps 2 and 6 exhaust the two possibilities according to whether \(G\setminus H\) contains an element of order \(2\). Therefore the number of elements of exact order \(2\) is either \(7\) or \(11\).