Prove that the intersection of any collection of normal subgroups is itself a normal
subgroup.

Question

Prove that the intersection of any collection of normal subgroups is itself a normal&amp;nbsp;subgroup.

RizerMix · Accepted Answer

To prove that the intersection of any collection of normal subgroups is itself a normal subgroup, we need to show that the intersection is a subgroup and that it is normal.
Let

G

be a group, and let

{H_{i}}_{i \in I}

be a collection of normal subgroups of

G

. We want to show that

K = ⋂_{i \in I} H_{i}

is a normal subgroup of

G

.
First, we show that

K

is a subgroup of

G

. Since each

H_{i}

is a subgroup of

G

, it follows that

K

is also a subgroup of

G

. To see this, note that

1 \in H_{i}

for each

i \in I

(since each

H_{i}

is a subgroup), and therefore

1 \in K

. Moreover, if

a, b \in K

, then

a, b \in H_{i}

for each

i \in I

, and hence

a b^{- 1} \in H_{i}

for each

i \in I

(since each

H_{i}

is a subgroup). Therefore,

a b^{- 1} \in K

. Thus,

K

is a subgroup of

G

.
Next, we show that

K

is a normal subgroup of

G

. To see this, let

g \in G

and let

k \in K

. We need to show that

g k g^{- 1} \in K

. Since

k \in K = ⋂_{i \in I} H_{i}

, we have

k \in H_{i}

for each

i \in I

. Since each

H_{i}

is normal, it follows that

g k g^{- 1} \in H_{i}

for each

i \in I

. Therefore,

g k g^{- 1} \in ⋂_{i \in I} H_{i} = K

. Thus,

K

is a normal subgroup of

G

.
Therefore, we have shown that the intersection of any collection of normal subgroups is itself a normal subgroup.

Prove that the intersection of any collection of

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