Question:
In the dihedral group \(D_5=\{r,s:\; r^5=e,\; s^2=e,\; sr=r^{-1}s\}\) under composition as the binary operation, which of the following options is NOT true?
In the dihedral group \(D_5=\{r,s:\; r^5=e,\; s^2=e,\; sr=r^{-1}s\}\) under composition as the binary operation, which of the following options is NOT true?
Show Hint
Check order, commutativity, center, and count all normal subgroups (including trivial and whole group) of D5.
Updated On: Jul 3, 2026
- \(D_5\) is a non-commutative group
- Order of \(D_5\) is 10
- \(Z(D_5)=\{e\}\)
- Number of normal subgroups of \(D_5\) is 2
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The Correct Option is D
Solution and Explanation
Step 1: List the elements of \(D_5\).
\(D_5=\{e,r,r^2,r^3,r^4,s,sr,sr^2,sr^3,sr^4\}\), so \(|D_5|=10\). This confirms option (B) is true.
\(D_5=\{e,r,r^2,r^3,r^4,s,sr,sr^2,sr^3,sr^4\}\), so \(|D_5|=10\). This confirms option (B) is true.
Step 2: Check commutativity.
From \(sr=r^{-1}s\) we get \(rs=sr^{-1}\ne sr\) (since \(r\ne r^{-1}\) for \(r\) of order 5). So \(D_5\) is non-abelian, confirming option (A) is true.
From \(sr=r^{-1}s\) we get \(rs=sr^{-1}\ne sr\) (since \(r\ne r^{-1}\) for \(r\) of order 5). So \(D_5\) is non-abelian, confirming option (A) is true.
Step 3: Find the center \(Z(D_5)\).
A rotation \(r^k\) (\(k\ne 0\)) commutes with \(s\) only if \(sr^k=r^ks\), but \(sr^k=r^{-k}s\), forcing \(r^{-k}=r^k\), i.e. \(r^{2k}=e\), i.e. \(5\mid 2k\), i.e. \(5\mid k\). For \(0<k<5\) this never happens, so no nontrivial rotation is central, and a similar check rules out reflections. Hence \(Z(D_5)=\{e\}\), confirming option (C) is true.
A rotation \(r^k\) (\(k\ne 0\)) commutes with \(s\) only if \(sr^k=r^ks\), but \(sr^k=r^{-k}s\), forcing \(r^{-k}=r^k\), i.e. \(r^{2k}=e\), i.e. \(5\mid 2k\), i.e. \(5\mid k\). For \(0<k<5\) this never happens, so no nontrivial rotation is central, and a similar check rules out reflections. Hence \(Z(D_5)=\{e\}\), confirming option (C) is true.
Step 4: Find all normal subgroups.
The subgroups of \(D_5\) are \(\{e\}\), the rotation subgroup \(\langle r\rangle=\{e,r,r^2,r^3,r^4\}\) of order 5, five subgroups of order 2 generated by each reflection \(\{e,sr^i\}\), and \(D_5\) itself.
The subgroups of \(D_5\) are \(\{e\}\), the rotation subgroup \(\langle r\rangle=\{e,r,r^2,r^3,r^4\}\) of order 5, five subgroups of order 2 generated by each reflection \(\{e,sr^i\}\), and \(D_5\) itself.
Step 5: Test normality of each.
\(\{e\}\) and \(D_5\) are always normal. \(\langle r\rangle\) has index 2, so it is normal. For a reflection subgroup \(\{e,sr^i\}\), conjugating by \(r\) gives \(r(sr^i)r^{-1}=sr^{-1}r^ir^{-1}=sr^{i-2}\) (using \(rs=sr^{-1}\)), which differs from \(sr^i\) since \(2\not\equiv 0\pmod 5\). So each reflection subgroup is conjugated to a different one; none of them is normal.
\(\{e\}\) and \(D_5\) are always normal. \(\langle r\rangle\) has index 2, so it is normal. For a reflection subgroup \(\{e,sr^i\}\), conjugating by \(r\) gives \(r(sr^i)r^{-1}=sr^{-1}r^ir^{-1}=sr^{i-2}\) (using \(rs=sr^{-1}\)), which differs from \(sr^i\) since \(2\not\equiv 0\pmod 5\). So each reflection subgroup is conjugated to a different one; none of them is normal.
Step 6: Count the normal subgroups.
The normal subgroups are exactly \(\{e\}\), \(\langle r\rangle\), and \(D_5\), a total of 3, not 2. So option (D) is false.
The normal subgroups are exactly \(\{e\}\), \(\langle r\rangle\), and \(D_5\), a total of 3, not 2. So option (D) is false.
Conclusion: Options (A), (B), (C) are true; option (D) is the one that is NOT true.
\[\boxed{\text{(D)}}\]
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