Problem 11

Question

Let $H_{1}$ be a subgroup of an abelian group $G_{1}$ and $H_{2}$ a subgroup of an abelian group $G_{2}$. Show that $H_{1} \times H_{2}$ is a subgroup of $G_{1} \times G_{2}$.

Step-by-Step Solution

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Answer

Question: Show that $H_1 \times H_2$ is a subgroup of $G_1 \times G_2$ if $H_1$ is a subgroup of $G_1$ and $H_2$ is a subgroup of $G_2$, where $G_1$ and $G_2$ are abelian groups. Answer: To prove that $H_1 \times H_2$ is a subgroup of $G_1 \times G_2$, we showed that the following conditions hold: the existence of an identity element in $H_1 \times H_2$, closure under the group operation, and the existence of inverses for all elements in $H_1 \times H_2$. Since all these conditions are satisfied, we can conclude that $H_1 \times H_2$ is a subgroup of $G_1 \times G_2$.

1Step 1: Identity element

First, let's find the identity element for the groups $H_1$ and $H_2$. Since $H_1$ is a subgroup of $G_1$, and $H_2$ is a subgroup of $G_2$, we know that the identity element of $H_1$ is also the identity element of $G_1$, denoted by $e_1$, and similarly, the identity element of $H_2$ is $e_2$. Now, let's verify that the identity element of $H_1 \times H_2$ is $(e_1, e_2)$. For any element $(h_1, h_2) \in H_1 \times H_2$, $(h_1, h_2)(e_1, e_2) = (h_1e_1, h_2e_2) = (h_1, h_2)$ and also $(e_1, e_2)(h_1, h_2) = (e_1h_1, e_2h_2) = (h_1, h_2)$. Since the identity element doesn't change the value of the element it's multiplied with, we can conclude that $(e_1, e_2)$ is the identity element of $H_1 \times H_2$.

2Step 2: Closure

Now we need to show that the product of any two elements in $H_1 \times H_2$ is also in $H_1 \times H_2$. Let $(h_1, h_2)$ and $(h'_1, h'_2)$ be any two elements in $H_1 \times H_2$. Then: $(h_1, h_2)(h'_1, h'_2) = (h_1h'_1, h_2h'_2)$. Since $H_1$ is a subgroup of $G_1$, and $H_2$ is a subgroup of $G_2$, the product $h_1h'_1$ is in $H_1$, and the product $h_2h'_2$ is in $H_2$. Therefore, the product $(h_1h'_1, h_2h'_2)$ is in $H_1 \times H_2$, proving closure.

3Step 3: Inverses

Lastly, we need to show that if $(h_1, h_2)$ is an element of $H_1 \times H_2$, then its inverse $(h_1^{-1}, h_2^{-1})$ is also in $H_1 \times H_2$. Since $H_1$ is a subgroup of $G_1$, we know that the inverse of $h_1$, denoted as $h_1^{-1}$, exists in $H_1$, and similarly, $h_2^{-1}$ exists in $H_2$. Therefore, the inverse $(h_1^{-1}, h_2^{-1})$ of the element $(h_1, h_2)$ exists in $H_1 \times H_2$. Since we have shown that the identity element exists, closure holds, and inverses exist for all elements in $H_1 \times H_2$, we can conclude that $H_1 \times H_2$ is a subgroup of $G_1 \times G_2$.

Problem 10

Problem 12

Other exercises in this chapter

Problem 9

Show that if $H$ is a subgroup of an abelian group $G,$ then a set $K \subseteq H$ is a subgroup of $G$ if and only if $K$ is a subgroup of $H$.

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Problem 10

Let $G$ be an abelian group with subgroups $H_{1}$ and $H_{2}$. Show that every subgroup $H$ of $G$ that contains $H_{1} \cup H_{2}$ must contain al

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Problem 12

Let $H_{1}$ be a subgroup of an abelian group $G_{1}$ and $H_{2}$ a subgroup of an abelian group $G_{2}$. Show that $H_{1} \times H_{2}$ is a subgroup

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Problem 13

Let $G_{1}$ and $G_{2}$ be abelian groups, and let $H$ be a subgroup of $G_{1} \times G_{2} .$ Define $$ H_{1}:=\left\\{a_{1} \in G_{1}:\left(a_{1}, a_{

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