Let $G_1$ be abelian group of order $6$ and $G_2=S_3$..For $j=1,2$, let $P_j$ be the statement: "$G_j$ has a unique subgroup of order $2$.then which of the following satement is true? a) both $P_1$ and $P_2$ are holds b)neither $P_1$ nor $P_2$ holds c)$P_1$ holds but $P_2$ not d)$P_2$ holds but $P_1$ not
Let G1 be abelian group of order 6 and G2=S3
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group-theory
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0What have you done so far? – 2017-01-01
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0$|G_1| = 2 \cdot 3$ is the product of two distinct primes and thus $G_1$ is cyclic. Also note that $\{\text{Id}, \, (1 \ 2) \}$ and $\{\text{Id}, \, (2 \ 3)\}$ are subgroups of $S_3$ of order 2. – 2017-01-01
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0thanks...sir I have an question..If a group is order of product of two distinct prime then it is cyclic?as for elxample 11 X13 =143..a group of order 143 is cyclic? – 2017-01-01
1 Answers
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$G_1 \simeq \Bbb Z_6 = \{0,1,2,3,4,5\}$. Here only subgroup of order $2$ is $\{0,3\}$.
Also $G_2=S_3=\{e,(1 \quad 2),(1 \quad 3),(2 \quad 3),(1 \quad 2 \quad 3),(1 \quad 3 \quad 2)\}$. Here $\{e,(1 \quad 2)\}$,$\{e,(2 \quad 3)\}$ and $\{e,(1 \quad 3)\}$ are the only subgroups of order $2$.